Answer:
Null hypothesis:[tex]\mu_{m} \geq \mu_{w}[/tex]
Alternative hypothesis:[tex]\mu_{m} < \mu_{w}[/tex]
[tex]p_v =P(t_{134}<-2.85)=0.0025[/tex]
D. Reject H0, there is sufficient evidence to conclude that the mean step pulse of men was less than the mean step pulse of women.
So on this case the 95% confidence interval would be given by [tex]-10.502 \leq \mu_{m} -\mu_w \leq -1.898[/tex]
We are 95% confident that the mean difference is in the confidence interval.
Step-by-step explanation:
1) Data given and notation
[tex]\bar X_{m}=112.5[/tex] represent the mean for the sample of men
[tex]\bar X_{w}=118.7[/tex] represent the mean for the sample women
[tex]s_{m}=11.1[/tex] represent the sample standard deviation for the sample men
[tex]s_{w}=14.2[/tex] represent the sample standard deviation for the sample eomen
[tex]n_{m}=56[/tex] sample size for the group men
[tex]n_{w}=80[/tex] sample size for the group women
z would represent the statistic (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:
Null hypothesis:[tex]\mu_{m} \geq \mu_{w}[/tex]
Alternative hypothesis:[tex]\mu_{m} < \mu_{w}[/tex]
We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:
[tex]t=\frac{\bar X_{m}-\bar X_{w}}{\sqrt{\frac{s^2_{m}}{n_{m}}+\frac{s^2_{w}}{n_{w}}}}[/tex] (1)
t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
3) Calculate the statistic
With the info given we can replace in formula (1) like this:
[tex]t=\frac{112.5-118.7}{\sqrt{\frac{11.1^2}{56}+\frac{14.2^2}{80}}}}=-2.85[/tex]
4) Statistical decision
The degrees of freedom are given by:
[tex]df=n_m +n_w -2= 56+80-2=134[/tex]
Since is a left tailed test the p value would be:
[tex]p_v =P(t_{134}<-2.85)=0.0025[/tex]
D. Reject H0, there is sufficient evidence to conclude that the mean step pulse of men was less than the mean step pulse of women.
5) Confidence interval
The confidence interval for the difference of means is given by the following formula:
[tex](\bar X_m -\bar X_w) \pm t_{\alpha/2}\sqrt{(\frac{s^2_m}{n_m}+\frac{s^2_w}{n_w})}[/tex] (1)
The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:
[tex]\bar X_m -\bar X_w =112.5-118.7=-6.2[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,134)".And we see that [tex]z_{\alpha/2}=1.98[/tex]
Now we have everything in order to replace into formula (1):
[tex]-6.2-1.98\sqrt{\frac{11.1^2}{56}+\frac{14.2^2}{80}}}=-10.502[/tex]
[tex]-6.2+1.98\sqrt{\frac{11.1^2}{56}+\frac{14.2^2}{80}}}=-1.898[/tex]
So on this case the 95% confidence interval would be given by [tex]-10.502 \leq \mu_{m} -\mu_w \leq -1.898[/tex]
We are 95% confident that the mean difference is in the confidence interval.
In the context of the study conducted by the physical therapist, the null hypothesis states that the mean step pulse of men equals the mean step pulse of women, and the alternative hypothesis states that the mean step pulse of men is less than that of women. With a t-test result that has a P-value less than 0.05, we reject the null hypothesis and accept the alternative hypothesis, indicating a significant difference between the mean step pulses of men and women.
Explanation:The physical therapist's study is essentially a hypothesis testing problem. The null hypothesis (σ_0) is that the mean step pulse of men is equal to the mean step pulse of women, and the alternative hypothesis (σ_a) is that the mean step pulse of men is less than the mean step pulse of women.
For this problem, we can write the hypotheses as follows:
σ_0: μ1 = μ2
σ_a: μ1 < μ2
The student's available choices seem to reflect this. Choosing between 'H0: μ1 = μ2; Ha :μ1>μ2' and 'H0:μ1 = μ2;Ha::μ1 not equal μ2'. The correct choice is not listed because the alternative hypothesis is incorrectly defined in both cases. The correct alternative hypothesis would be 'μ1 < μ2', pointing that the mean step pulse of men is less than that of women.
The t-test result given in the question 'T = 2.85 P = 0.0025 DF = 132' indicates that the mean step pulse of men is not equal to that of women, with the P-value being less than 0.05. Thus, with this information, you would typically reject the null hypothesis and accept the alternative hypothesis, which suggests a significant difference between the mean step pulses of men and women.
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Prove that two right triangles are congruent if the corresponding altitudes and angle bisectors through the right angles are congruent.
Two right triangles are congruent if the corresponding altitudes and angle bisectors through the right angles are congruent.
Explanation:In order to prove that two right triangles are congruent if the corresponding altitudes and angle bisectors through the right angles are congruent, we can use the side-side-side (SSS) congruence theorem. This theorem states that if three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent.
In this case, we can show that the corresponding altitudes and angle bisectors through the right angles are congruent for both triangles. Since both triangles have congruent corresponding altitudes and congruent angle bisectors through the right angles, we can conclude that the triangles are congruent.
Therefore, the statement is proven.
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A consumer activist group wants to determine the mean lifetime of the Amazon Kindle DX. The consumer activist groups randomly selects 25 Kindles and finds that the average lifespan was 38 months with standard deviation 12 months. Find a 95% confidence interval for the population mean lifetime of the Amazon Kindle DX.
Answer:
Confidence interval for the population mean lifetime of the Amazon Kindle DX is (33.30 months to 42.70 months)
Step-by-step explanation:
Given;
Mean lifespan x = 38 months
Standard deviation r = 12 months
Number of kindle selected n = 25
Confidence range = 95%
Z*(95%) = 1.96
Confidence interval = x+/-Z*(r/√n)
= 38 +/- 1.96(12/√25)
= 38 +/- 4.70
Confidence interval = (33.30 months to 42.70 months)
a chemist needs 120 milliliters of a 72% solution but has only 51% and 87% solutions available. Find how many milliliters of each that should be mixed to get the desired solution.
Answer:
The chemist needs 50 mL of 51% solution and 70 mL of 87% solution.
Step-by-step explanation:
If x is the volume of 51% solution, and y is the volume of 87% solution, then:
x + y = 120
0.51x + 0.87y = 0.72(120)
Solve the system of equations.
0.51x + 0.87(120 − x) = 0.72(120)
0.51x + 104.4 − 0.87x = 86.4
0.36x = 18
x = 50
y = 70
The chemist needs 50 mL of 51% solution and 70 mL of 87% solution.
The incorrect work of a student to solve an equation 2(y + 8) = 4y is shown below:
Step 1: 2(y + 8) = 4y
Step 2: 2y + 10 = 4y
Step 3: 2y = 10
Step 4: y = 5
Which of the following explains how to correct Step 2 and shows the correct value of y? (5 points)
Group of answer choices
Answer:
Step 2 involved distributive property and the value of y is equal to 8.
Step-by-step explanation:
In step 2, there has been an error in applying the Distributive Property correctly.
Distributive Property- a(b+c)
= a x b + a x c
Step 2: [tex]2y+16=4y[/tex]
Step 3:[tex]2y=16[/tex]
Step 4: [tex]y=8[/tex]
y=8
Answer:
2 should be distributed as 2y + 16; y + 8
ureg placed 32 chairs in the auditorium. There are 8 chairs in each row. Which
equation could be used to represent this situation
A. 32 x 8
00
--
B. 32 + 8
00
8 = 32
D.
00
x 8 = 32
The equation to represent this situation is 8x = 32
Solution:
Given that ureg placed 32 chairs in the auditorium
There are 8 chairs in each row
To find: equation used to represent this situation
From given information,
1 row = 8 chairs
So let us find an expression to determine the number of rows to place 32 chairs
Let "x" be the number of rows required to place 32 chairs
Since 1 row contains 8 chairs, expression to determine the number of rows to place 32 chairs is given as:
[tex]8 \times \text{ number of row } = 32[/tex]
8x = 32
Thus the equation to represent this situation is 8x = 32
The lifetime of a certain type of battery is normally distributed with mean value 15 hours and standard deviation 1 hour. There are four batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages? (Round your answer to two decimal places.)
Answer:
If the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 15
Standard Deviation, σ = 1
Sample size = 4
Total lifetime of 4 batteries = 40 hours
We are given that the distribution of lifetime is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Standard error due to sampling:
[tex]\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{1}{\sqrt4} = 0.5[/tex]
We have to find the value of x such that the probability is 0.05
P(X > x) = 0.05
[tex]P( X > x) = P( z > \displaystyle\frac{x - 40}{0.5})=0.05[/tex]
[tex]= 1 -P( z \leq \displaystyle\frac{x - 40}{0.5})=0.05 [/tex]
[tex]=P( z \leq \displaystyle\frac{x - 40}{0.5})=0.95 [/tex]
Calculation the value from standard normal z table, we have,
[tex]\displaystyle\frac{x - 40}{0.5} = 1.64\\x = 40.825 \approx 40.83[/tex]
Hence, if the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.
Students in a discussion of gun control in a sociology class at Foothill Community College argue that Republicans are more likely to oppose gun control than Independents. They use data from an article titled "Gun Control Splits America," published March 23, 2010 in pewresarch.org by the Pew Research Center for the People and the Press. In this study 62% of Republicans and 57% of Independents say that states should not be able to pass laws banning handguns.
For a claim that a larger proportion of Republicans oppose state laws banning handguns when compared to Independents, the null and alternative hypotheses are
H0: p1-p2 = 0 (p1 = p2)
Ha: p1-p2 > 0 (p1 > p2)
The p -value is 0.06. If we conduct this test at a 5% level of significance, what would be an appropriate conclusion?
A. Reject H0 , and support Ha.
B. Support H0 , and reject Ha.
C. Fail to Reject H0.
D. do not support Ha .
Answer:
C. Fail to Reject H0.
Step-by-step explanation:
If the P-value is 0.06, that means that the result enters in the acceptance region. It is a value that is expected to happen if both proportions are equal (null hypothesis) at this significance level.
The conclusion when the P-value is bigger than the significance level is that the effect is not significant and it failed to reject the null hypothesis.
Final answer:
The null and alternative hypotheses for the claim that a larger proportion of Republicans oppose state laws banning handguns when compared to Independents are H0: p1-p2 = 0 (p1 = p2) and Ha: p1-p2 > 0 (p1 > p2).
The p-value of 0.06 is larger than the significance level of 0.05, so the appropriate conclusion is C. Fail to Reject H0.
Explanation:
The null and alternative hypotheses for the claim that a larger proportion of Republicans oppose state laws banning handguns when compared to Independents are:
H0: p1-p2 = 0 (p1 = p2)
Ha: p1-p2 > 0 (p1 > p2)
The p-value of 0.06 is larger than the significance level of 0.05. Therefore, we fail to reject the null hypothesis. Hence, an appropriate conclusion is:
C. Fail to Reject H0.
Determine the horizontal change of a line with an x intercept at (3,0) and a y intercept at (0,2)
Answer:the horizontal change of the line is 3
Step-by-step explanation:
The horizontal change is the change in the value of x on the horizontal axis. It is expressed as
x2 - x1
Where
x2 represents the final value of x
x1 represents the initial value of x
An x intercept at (3,0) means that the line cut across the x axis at the point when x = 3 and y = 0
A y intercept at (0,2) means that the line cut across the y axis at the point when 0 = 3 and y = 2
Change in the horizontal axis would be x2 - x1 = 3 - 0 = 3
Researchers continue to find evidence that brains of adolescents behave quite differently than either brains of adults or brains of children. In particular, adolescents seem to hold on more strongly to fear associations than either children or adults, suggesting that frightening connections made during the teen years are particularly hard to unlearn. In one study,1 participants first learned to associate fear with a particular sound. In the second part of the study, participants heard the sound without the fear-causing mechanism, and their ability to "unlearn" the connection was measured. A physiological measure of fear was used, and larger numbers indicate less fear. We are estimating the difference in mean response between adults and teenagers. The mean response for adults in the study was 0.225 and the mean response for teenagers in the study was 0.059. We are told that the standard error of the estimate is 0.091. Let group 1 be adults and group 2 be teenagers.
(a) Give notation for the quantity that is being estimated.
Answer:
a) [tex]\mu_1 -\mu_2[/tex] parameter of interest.
Where [tex]\mu_1[/tex] represent the mean response for adults
[tex]\mu_2[/tex] represent the mean response for teenegers
b) The best estimate is given by [tex]\bar X_1 -\bar X_2[/tex]
Since the best estimator for the true mean is the sample mean [tex]\hat \mu = \bar X[/tex]
c) The best estimate is given by [tex]\bar X_1 -\bar X_2 =0.225-0.059=0.166[/tex]
d) The 95% confidence interval would be given by [tex]-0.012 \leq \mu_1 -\mu_2 \leq 0.344[/tex]
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Let group 1 be adults and group 2 be teenagers.
[tex]\bar X_1 =0.225[/tex] represent the sample mean 1
[tex]\bar X_2 =0.059[/tex] represent the sample mean 2
n1 represent the sample 1 size
n2 represent the sample 2 size
[tex]s_1 [/tex] sample standard deviation for sample 1
[tex]s_2 [/tex] sample standard deviation for sample 2
SE =0.091 represent the standard error for the estimate
(a) Give notation for the quantity that is being estimated.
[tex]\mu_1 -\mu_2[/tex] parameter of interest.
(b) Give notation for the quantity that gives the best estimate.
[tex]\mu_1 -\mu_2[/tex] parameter of interest.
The best estimate is given by [tex]\bar X_1 -\bar X_2[/tex]
Since the best estimator for the true mean is the sample mean [tex]\hat \mu = \bar X[/tex]
(c) Give the value for the quantity that gives the best estimate.
The best estimate is given by [tex]\bar X_1 -\bar X_2 =0.225-0.059=0.166[/tex]
(d) Give a confidence interval for the quantity being estimated. Assuming 95% of confidence
The confidence interval for the difference of means is given by the following formula:
[tex](\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}[/tex] (1)
The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:
[tex]\bar X_1 -\bar X_2 =0.225-0.059=0.166[/tex]
We can assume that since we know the standard error the deviations are known and we can use the z distribution instead of the t distribution for the confidence interval.
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that [tex]z_{\alpha/2}=1.96[/tex]
The standard error is given by the following formula:
[tex]SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}=0.091[/tex]
Given by the problem
Now we have everything in order to replace into formula (1):
[tex]0.166-1.96(0.091)=-0.012[/tex]
[tex]0.166+1.96(0.091)=0.344[/tex]
So on this case the 95% confidence interval would be given by [tex]-0.012 \leq \mu_1 -\mu_2 \leq 0.344[/tex]
Final answer:
The quantity being estimated in the study is the difference in mean response to unlearn fear associations between adults and teenagers, denoted by Δμ = μ1 - μ2, where μ1 and μ2 represent the mean responses for adults and teenagers, respectively. This study contributes to understanding how fear associations are formed and unlearned, with implications on evolutionary predisposition towards certain fears.
Explanation:
The quantity being estimated in the study between adolescents and adults regarding their ability to unlearn fear associations tied to a specific sound is captured by the notation Δμ = μ1 - μ2. Here, μ1 represents the mean response for adults, and μ2 represents the mean response for teenagers. In this context, a higher physiological measure indicates less fear, with adults showing a mean response of 0.225 and teenagers showing a mean response of 0.059. The standard error of the estimate provided is 0.091, which helps in understanding the variability or precision of our estimated difference between the two groups' mean responses.
This study hints at the broader theory of preparedness, suggesting that humans are evolutionarily predisposed to easily associate certain stimuli with fear. Notably, the differentiation in fear response unlearning between age groups aligns with observations in social and developmental psychology about the specificity of fear acquisition and the challenges in modify these responses once established, especially during the teenage years.
For each of the given situations, write out the alternative hypothesis, being sure to state whether it is one-sided or two-sided.a) A consumer magazine discovered that 30% of a certain computer model had warranty problems over the first three months. From a random sample, the manufacturer wants to know if a new model has improved that rate.Complete the alternative hypothesis and determine whether the alternative hypothesis is one-sided or two-sided.
Answer:
Null hypothesis: [tex]p\leq 0.3[/tex]
Alternative hypothesis: [tex]p > 0.3[/tex]
Step-by-step explanation:
1) Previous concepts
A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".
The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".
The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".
2) Solution to the problem
On this case we want to test is [tex]p>0.3[/tex] since we want to check if the new model has improved the warranty rate, we can express it like this:
[tex]p-0.3<0[/tex] since are equivalent expressions.
And the alternative hypothesis should be the complement:
Null hypothesis: [tex]p\leq 0.3[/tex] or [tex]p=0.3[/tex]
So the correct system of hypothesis for this case would be:
Null hypothesis: [tex]p\leq 0.3[/tex]
Alternative hypothesis: [tex]p > 0.3[/tex]
The alternative way should be:
Null hypothesis: [tex]p = 0.3[/tex]
Alternative hypothesis: [tex]p > 0.3[/tex]
Antonette gets $70\%$ on a 10-problem test, $80\%$ on a 20-problem test and $90\%$ on a 30-problem test. If the three tests are combined into one 60-problem test, which percent is her overall score, rounded to the nearest percent?
Answer:
Percentage score will be 83.33 %
Step-by-step explanation:
We have given Antonette gets 70 % on 10 problem test
Let consider here here total problem = total marks
So marks get 10 10 problem test = 10×0.7 = 7
Marks get in 20 problem test = 20×0.8 = 16
And marks get in 30 problem test = 30×0.9 = 27
Now total marks get get by Antonette = 7 +16 + 27 = 50
And total marks = 60
So percentage score of Antonette [tex]=\frac{50}{60}\times 100=83.33[/tex] %
All fifth-grade students are given a test on academic achievement in New York State. Suppose the mean score is 70 for the entire state. A random sample of fifth-grade students is selected from Long Island. Below are the scores in this sample from a normal population. 82 94 66 87 68 85 68 84 70 83 65 70 83 71 82 72 73 81 76 74 a. Construct a 95% confidence interval for the population mean score on Long Island. b. Construct a 90% confidence interval for the population standard deviation of the scores on Long Island. c. A teacher at a Long Island high school claims that the mean score on Long Island is higher than the mean for New York State. Conduct a test to see if this claim is reasonable using α = 0.01. d. Find the p-value of the test.
Answer:
a) The 90% confidence interval would be given by (73.56;79.84)
b) The 90% confidence interval for the deviation would be [tex] 6.44 \leq \sigma \leq 11.116[/tex].
c) If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than 70 at 1% of significance. So the claim of the teacher makes sense.
d) Since is a one-side upper test the p value would given by:
[tex]p_v =P(t_{19}>3.69)=0.00078[/tex]
Step-by-step explanation:
Data given: 82 94 66 87 68 85 68 84 70 83 65 70 83 71 82 72 73 81 76 74
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
We can calculate the sample mean and deviation with the following formulas:
[tex]\bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
[tex]s=\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]
And we got the following results:
[tex]\bar X=76.7[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]s=8.112[/tex] represent the population standard deviation
n=20 represent the sample size
90% confidence interval
Part a
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The degrees of freedom are given by:
[tex]df=n-1=20-1=19[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,19)".And we see that [tex]t_{\alpha/2}=1.73[/tex]
Now we have everything in order to replace into formula (1):
[tex]76.7-1.73\frac{8.112}{\sqrt{20}}=73.56[/tex]
[tex]76.7+1.73\frac{8.112}{\sqrt{20}}=79.84[/tex]
So on this case the 90% confidence interval would be given by (73.56;79.84)
Part b
The confidence interval for the population variance is given by the following formula:
[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]
On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.
The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:
[tex]df=n-1=20-1=19[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical values.
The excel commands would be: "=CHISQ.INV(0.05,19)" "=CHISQ.INV(0.95,19)". so for this case the critical values are:
[tex]\chi^2_{\alpha/2}=30.143[/tex]
[tex]\chi^2_{1- \alpha/2}=10.117[/tex]
And replacing into the formula for the interval we got:
[tex]\frac{(19)(8.112^2)}{30.143} \leq \sigma^2 \leq \frac{(19)(8.112^2)}{10.117}[/tex]
[tex] 41.472 \leq \sigma^2 \leq 123.574[/tex]
So the 90% confidence interval for the deviation would be [tex] 6.44 \leq \sigma \leq 11.116[/tex].
Part c
Null hypothesis:[tex]\mu \leq 70[/tex]
Alternative hypothesis:[tex]\mu > 70[/tex]
Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{76.7-70}{\frac{8.112}{\sqrt{20}}}=3.69[/tex]
Part d
P-value
Since is a one-side upper test the p value would given by:
[tex]p_v =P(t_{19}>3.69)=0.00078[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than 70 at 1% of significance.
Answer:
(a) 95% confidence interval for the population mean score on Long Island is (73, 80.40)
(b) 90% confidence interval for the population standard deviation of the scores on Long Island is (4.85, 10.97)
(c) The teacher's claim that the mean score on Long Island is higher than the mean for New York State is reasonable
(d) p-value is 0.01
Step-by-step explanation:
From the data values from Long Island,
Mean is 76.7 and standard deviation is 7.91
Confidence Interval (CI) = mean + or - (t × sd)/√n
(a) mean = 76.7, sd = 7.91, n = 20, degree of freedom = n-1 = 20-1 = 19, t-value corresponding to 19 degrees of freedom and 95% confidence level is 2.093
Lower bound = 76.7 - (2.093×7.91)/√20 = 76.7 - 3.70 = 73
Upper bound = 76.7 + (2.093×7.91)/√20 = 76.7 + 3.70 = 80.40
95% CI is (73, 80.40)
(b) CI = sd + or - (t×sd)/√n
sd = 7.91, t-value corresponding to 19 degrees of freedom and 90% confidence level is 1.729
Lower bound = 7.91 - (1.729×7.91)/√20 = 7.91 - 3.06 = 4.85
Upper bound = 7.91 + (1.729×7.91)/√20 = 7.91 + 3.06 = 10.97
90% CI is (4.85, 10.97)
(c) Null hypothesis: The mean score on Long Island is 70
Alternate hypothesis: The mean score on Long Island is greater than 70
Z = (sample mean - population mean)/(sd/√n) = (70 - 76.7)/(7.91/√20) = -6.7/1.77 = -3.79
Using 0.01 significance level, the critical value is 2.326
Since -3.79 is less than 2.326, reject the null hypothesis. The teacher's claim is reasonable
(d) p-value = 1 - cumulative area of test statistic = 1 - 0.9900 = 0.01
Weatherwise is a magazine published by the American Meteorological Society. One issue gives a rating system used to classify Nor'easter storms that frequently hit New England and can cause much damage near the ocean. A severe storm has an average peak wave height of μ = 16.4 feet for waves hitting the shore. Suppose that a Nor'easter is in progress at the severe storm class rating. Peak wave heights are usually measured from land (using binoculars) off fixed cement piers. Suppose that a reading of 38 waves showed an average wave height of x= 17.3 feet. Previous studies of severe storms indicate that σ = 3.3 feet. Does this information suggest that the storm is (perhaps temporarily) increasing above the severe rating? Use α = 0.01. Solve the problem using the critical region method of testing (i.e., traditional method). (Round your answers to two decimal places.)test statistic = critical value = State your conclusion in the context of the application.Reject the null hypothesis, there is sufficient evidence that the average storm level is increasing.Reject the null hypothesis, there is insufficient evidence that the average storm level is increasing. Fail to reject the null hypothesis, there is sufficient evidence that the average storm level is increasing.Fail to reject the null hypothesis, there is insufficient evidence that the average storm level is increasing.Compare your conclusion with the conclusion obtained by using the P-value method. Are they the same?
Answer:
Step-by-step explanation:
A psychiatrist studying the affects of healthy eating habits on mood improvement. He identifies 25 people who eat healthy food and 25 who do not. Each other 50 people is given a questionnaire design to determine their mood. None of the 50 people participated in the study knew they were part of the study. Which statement is true? A this is a random mise comparative experiment be this is a double blind study see this is an observational study D this is matched pairs design
Answer:
The statement which is true is:
B. This is a double blind study
Step-by-step explanation:
Double Blind Study is such a study method in which participants don't know about the information that can effect the participants in order to eliminate the bias. This is the case with out situation in which participants didn't know that they were the part of a study. Randomized Comparative Experiment is such an experiment in which participants are randomly selected for different treatments as well as the comparison of effects of different treatments is done. So, the option A is not valid in our situation.Observational Study is such a study in which we only observe the the data collected or participants. We observe the cause-effect relationship in this study. Matched Pair Design is such an experiment in which subjects are grouped in pairs. It is a special type of randomized block design.Answer:
C. Observational Study
Step-by-step explanation:
There arent any treatments being manipulated so it can't be an experiment. The researcher is aware of the experiment being performed so its not double-blind. The only option it could be is C, because the researcher is observing the people eating and isn't manipulating their eating habits in any way.
Ray Flagg took out a 60-month fixed installment loan of $12,000 to open a new pet store. He paid no money down and began making monthly payments of $232. Ray's business does better than expected and instead of making his 30th payment, Ray wishes to repay his loan in full.
Answer:
Ray Flagg will pay $5,272 at the time of his 30th installment.
Step-by-step explanation:
Ray took $12,000 load for 60 months. As he paid no amount as down payment so his monthly payment will be $200:
[tex]=12000/60\\=200[/tex]
Instead of $200 per month, he used to pay $232 per month. So, before his 30th installment, he paid 29 installments each of $232 which is $6,728:
[tex]=232*29\\=6728[/tex]
As the business does better, he wishes to payback remaining amount at once so he will pay $5,272 as:
[tex]12000-6728\\=5272[/tex]
To calculate the remaining balance of a fixed installment loan after a certain number of payments, use the formula provided in the detailed answer.
Explanation:Mathematics: High SchoolRay Flagg took out a 60-month fixed installment loan of $12,000 to open a new pet store. He paid no money down and began making monthly payments of $232. Ray's business does better than expected and instead of making his 30th payment, Ray wishes to repay his loan in full.
To calculate the remaining balance after 29 months, we can use the formula for the remaining balance of a fixed installment loan:
Remaining Balance = Balance × (1 + Monthly Interest Rate)Number of Payments Made - (Monthly Payment × ((1 + Monthly Interest Rate)Number of Payments Made - 1) / Monthly Interest Rate)
Using the given values, the monthly interest rate can be calculated by dividing the annual interest rate by 12 and converting it to a decimal.
Finally, substitute the values into the formula to find the remaining balance after 29 months.
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7. Solving for dominant strategies and the Nash equilibrium Suppose Nick and Rosa are playing a game in which both must simultaneously choose the action Left or Right. The payoff matrix that follows shows the payoff each person will earn as a function of both of their choices. For example, the lower-right cell shows that if Nick chooses Right and Rosa chooses Right, Nick will receive a payoff of 6 and Rosa will receive a payoff of 5. Rosa Left Right Nick Left 8, 4 4, 5 Right 5, 4 6, 5
In a game of choice and payoff, Nick's dominant strategy is to choose 'Right', while Rosa lacks a dominant strategy. The Nash equilibrium is when Nick chooses 'Right' and Rosa chooses either 'Left' or 'Right' because changing their decisions would not lead to higher payoff.
Explanation:The subject of this question is about a concept from game theory known as dominant strategies and the Nash equilibrium. Nick and Rosa are playing a game where they each simultaneously choose an action (Left or Right) and receive a payoff that depends on both their choices. To find the dominant strategy for each player, we need to identify what action that player would take, regardless of the other player's choice.
For Nick, the dominant strategy is to choose Right because his payoff (5 when Rosa picks Left, 6 when Rosa picks Right) is higher than when he picks Left (8 when Rosa picks Left, 4 when Rosa picks Right). For Rosa, she has no dominant strategy because her payoff is the same (4) whether she chooses Left or Right if Nick is choosing Left, and the same holds if Nick is choosing Right.
The Nash equilibrium is a situation where neither player can benefit by changing their strategy, assuming the other player stays the same. Here, the Nash equilibrium occurs when Nick chooses Right and Rosa chooses Left or Right, because neither player can gain a higher payoff by unilaterally changing their strategy.
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The average maximum monthly temperature in a city is 29.9 degrees Celsius. The standard deviation in maximum monthly temperature is 2.31 degrees. Assume that maximum monthly temperatures are normally distributed. Use this Rule of Thumb to complete the sentence. Round your answers to one decimal place. (Enter your answers from smallest to largest.) 95% of the time the maximum monthly temperature is between and degrees Celsius.
Answer: 95% of the time the maximum monthly temperature is between 25.28 degrees Celsius and 34.52 degrees Celsius.
Step-by-step explanation:
The Range Rule of Thumb tells that the range is approximately four times the standard deviation.
95% of the data lies within 2 standard deviations from the mean .
Maximum usual value = Mean +2 (Standard deviation )
Minimum usual value = Mean - 2 (Standard deviation)
Given : Mean = 29.9 degrees Celsius
Standard deviation = 2.31 degrees Celsius
Then, according to the Range Rule of Thumb , we have
Maximum usual value = 29.9 +2 (2.31) = 34.52 degrees Celsius
Minimum usual value = 29.9 - 2 (2.31) = 25.28 degrees Celsius
i.e. 95% of the time the maximum monthly temperature is between 25.28 degrees Celsius and 34.52 degrees Celsius.
Using the empirical rule for normal distribution, 95% of the time, the maximum monthly temperature is between 25.3 and 34.5 degrees Celsius, given the mean is 29.9°C and the standard deviation is 2.31°C.
Explanation:The question asks for the range of temperatures within which 95% of the maximum monthly temperatures fall, given that the average maximum monthly temperature is 29.9 degrees Celsius with a standard deviation of 2.31 degrees, assuming a normal distribution. To find this, we can apply the empirical rule that states approximately 95% of the data in a normal distribution falls within two standard deviations of the mean. Therefore:
Calculate the lower boundary by subtracting two standard deviations from the mean: 29.9 - (2 × 2.31) = 29.9 - 4.62 = 25.3 degrees Celsius.Calculate the upper boundary by adding two standard deviations to the mean: 29.9 + (2 × 2.31) = 29.9 + 4.62 = 34.5 degrees Celsius.Thus, 95% of the time, the maximum monthly temperature is between 25.3 and 34.5 degrees Celsius.
In order to comply with the Environmental protection Agency (EPA) regulations of the Clean Water Act, a large agricultural company wants to know the average nitrogen concentration in the soil of an agricultural region it plans to purchase. The seller claims that the average nitrogen level does not exceed 0.49 units. To test this claim at 0.05 level of significance, nitrogen concentration of soil samples were recorded at 51 sites in that agricultural region. The sample mean was found to be 0.505 and the sample standard deviation 0.12.
Answer:
[tex]t=\frac{0.505-0.49}{\frac{0.12}{\sqrt{51}}}=0.893[/tex]
[tex]p_v =P(t_{50}>0.893)=0.1881[/tex]
If we compare the p value and a significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean is not significantly higher than 0.49 units.
Step-by-step explanation:
Data given and notation
[tex]\bar X=0.505[/tex] represent the sample mean
[tex]s=0.12[/tex] represent the standard deviation for the sample
[tex]n=51[/tex] sample size
[tex]\mu_o =0.49[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses to be tested
We need to conduct a hypothesis in order to determine if the average nitrogen level dos not exced 0.49 units, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 0.49[/tex]
Alternative hypothesis:[tex]\mu > 0.49[/tex]
Compute the test statistic
We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
We can replace in formula (1) the info given like this:
[tex]t=\frac{0.505-0.49}{\frac{0.12}{\sqrt{51}}}=0.893[/tex]
Now we need to find the degrees of freedom for the t distirbution given by:
[tex]df=n-1=51-1=50[/tex]
What do you conclude?
Compute the p-value
Since is a right tailed test the p value would be:
[tex]p_v =P(t_{50}>0.893)=0.1881[/tex]
If we compare the p value and a significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean is not significantly higher than 0.49 units.
A telemetry voltage V, transmitted from a position sensor on a ship's rudder, is a random variable with PDF:
fV(v)={1/32 0 −16<_v<_16, otherwise.
A receiver in the ship's control room receive R = V+X. The random variable X is a Gaussian (4,4) noise voltage that is independent of V. The receiver uses R to calculate a linear estimate of the telemetry voltage: V = aR+b.
(a) Find the expected value of the received voltage. E [R] = _______
(b) Find the variance of the received voltage. Var [R] = _________
(c) Find the covariance of the transmitted and received voltage. Cov [V, R] = _________
(d) Find the optimal linear estimate. VL (R) = __________
(e) Compute the minimum mean square error of the estimate. e* = __________
The expected value of the received voltage is 4. The variance of the received voltage is 64/3. The covariance of the transmitted and received voltage is 0.
(a) To find the expected value of the received voltage, we need to use the linearity property of the expectation and the fact that V and X are independent. The expected value of R is given by:
E[R] = E[V+X] = E[V] + E[X]
Since V and X are independent, we have E[X] = 4 and E[V] = 0 (by symmetry of the uniform distribution). Therefore, E[R] = 0 + 4 = 4.
(b) To find the variance of the received voltage, we can use the properties of variance. Variance is additive for independent random variables, so:
Var[R] = Var[V+X] = Var[V] + Var[X]
Since V and X are independent, we have Var[X] = 4^2 = 16 and Var[V] = (16^2)/12 = 64/12 = 16/3. Therefore, Var[R] = 16/3 + 16 = 64/3.
(c) The covariance of the transmitted and received voltage is given by:
Cov[V, R] = E[(V - E[V])(R - E[R])]
Since E[V] = 0 and E[R] = 4, this simplifies to:
Cov[V, R] = E[VR] - E[V]E[R]
Since V and R are independent, we have Cov[V, R] = E[V]E[R] - E[V]E[R] = 0.
(d) The optimal linear estimate VL(R) is given by:
VL(R) = E[V] + Cov[V, R]/Var[R] * (R - E[R])
Since Cov[V, R] = 0, the optimal linear estimate becomes:
VL(R) = E[V] + 0/Var[R] * (R - E[R]) = E[V] = 0.
(e) The minimum mean square error of the estimate is given by:
e* = Var[V - VL(R)]
Since VL(R) = E[V] = 0, this simplifies to:
e* = Var[V] = 16/3.
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there are 24 panes of glass in 8 windows. use ratio to complete the table below
Answer: It's attached.
Step-by-step explanation:
The table is attached.
The ratio is:
[tex]ratio=\frac{24}{8}\\\\ratio=3[/tex]
Knowing tha ratio, you can complete the table.
The steps are:
1. Multiply the number of panes given in the table by the ratio find above, in order to find the number of windows.
3. Divide the number of windows given in the table by the ratio find above, in order to find the number of panes.
Given [tex]Panes=3[/tex]:
[tex]Windows=3*3=9[/tex]
Given [tex]Windows=3[/tex]:
[tex]Panes=\frac{3}{3}=1[/tex]
Given [tex]Windows=5[/tex]:
[tex]Panes=\frac{5}{3}[/tex]
Given [tex]Panes=18[/tex]:
[tex]Windows=18*3=54[/tex]
Answer:windows x6
Step-by-step explanation:
State the half-angle identities used to integrate sin^(2) x and cos^(2) x
the half-angle formulas are sin ^(2)x = ?? and cos ^(2)x = ??
Answer:
the answer is D
Step-by-step explanation:
Think of the Pythagorean Theorem which states that a^2 + b^2 = c^2. The Pythagorean Identities used in trigonometry are the angle version which can be used to simplify expressions.
HOPE THIS HELPED ;3
The formula of half angle identities is sin²x = (1-cos(x/2))/2 and cos²x = (1+cos(x/2))/2.
What is angle?An angle is the formed when two straight lines meet at one point, it is denoted by θ.
The given terms are,
sin²x and cos²x
To integrate sin²x, the formula is used,
sin²x = (1-cos(x/2))/2
To integrate cos²x, the formula is used,
cos²x = (1+cos(x/2))/2
The formula is, sin²x = (1-cos(x/2))/2 and cos²x = (1+cos(x/2))/2.
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For a given piece of code, the hit ratio of first cache is 0.1 and the hit ratio of second cache is 0.3. The time required to access the first cache is 10 nanoseconds, the second cache is 100 nanoseconds, and the time to access the underlying physical memory is 1 microsecond, what is the effective memory access time for the piece of code?
Answer:
effective memory access = 658 ns
Step-by-step explanation:
GIven data:
Effective memory access time is given as
[tex] = [H_1*T_1]+[(1-H_1)*H_2*T_2]+[(1-H_1)(1-H_2)*H_m*T_m][/tex]
from the data given above we have
[tex]H_1 = 0.1[/tex]
[tex]H_2 = 0.3[/tex]
[tex]T_1 = 10 ns[/tex]
[tex]T_2 = 100 ns[/tex]
hit rate, [tex]H_m = 1 ns[/tex]
access time [tex]= T_m = 1000 ns[/tex]
Plugging all information in above formula to get the effective memory access
[tex]= 0.1\times 10 + 0.9\times 100+ 0.9 \times 0.7\times 1 \times 1000[/tex]
= 1+27+ 630
=658 ns
Consider the integral Integral from 0 to 1 e Superscript 6 x Baseline dx with nequals 25 . a. Find the trapezoid rule approximations to the integral using n and 2n subintervals. b. Find the Simpson's rule approximation to the integral using 2n subintervals. c. Compute the absolute errors in the trapezoid rule and Simpson's rule with 2n subintervals.
Answer:
a.
With n = 25, [tex]\int_{0}^{1}e^{6 x}\ dx \approx 67.3930999748549[/tex]
With n = 50, [tex]\int_{0}^{1}e^{6 x}\ dx \approx 67.1519320308594[/tex]
b. [tex]\int_{0}^{1}e^{6 x}\ dx \approx 67.0715427161943[/tex]
c.
The absolute error in the trapezoid rule is 0.08047
The absolute error in the Simpson's rule is 0.00008
Step-by-step explanation:
a. To approximate the integral [tex]\int_{0}^{1}e^{6 x}\ dx[/tex] using n = 25 with the trapezoid rule you must:
The trapezoidal rule states that
[tex]\int_{a}^{b}f(x)dx\approx\frac{\Delta{x}}{2}\left(f(x_0)+2f(x_1)+2f(x_2)+...+2f(x_{n-1})+f(x_n)\right)[/tex]
where [tex]\Delta{x}=\frac{b-a}{n}[/tex]
We have that a = 0, b = 1, n = 25.
Therefore,
[tex]\Delta{x}=\frac{1-0}{25}=\frac{1}{25}[/tex]
We need to divide the interval [0,1] into n = 25 sub-intervals of length [tex]\Delta{x}=\frac{1}{25}[/tex], with the following endpoints:
[tex]a=0, \frac{1}{25}, \frac{2}{25},...,\frac{23}{25}, \frac{24}{25}, 1=b[/tex]
Now, we just evaluate the function at these endpoints:
[tex]f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1[/tex]
[tex]2f\left(x_{1}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281[/tex]
[tex]2f\left(x_{2}\right)=2f\left(\frac{2}{25}\right)=2 e^{\frac{12}{25}}=3.23214880438579[/tex]
...
[tex]2f\left(x_{24}\right)=2f\left(\frac{24}{25}\right)=2 e^{\frac{144}{25}}=634.696657835701[/tex]
[tex]f\left(x_{25}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735[/tex]
Applying the trapezoid rule formula we get
[tex]\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{50}(1+2.54249830064281+3.23214880438579+...+634.696657835701+403.428793492735)\approx 67.3930999748549[/tex]
To approximate the integral [tex]\int_{0}^{1}e^{6 x}\ dx[/tex] using n = 50 with the trapezoid rule you must:We have that a = 0, b = 1, n = 50.
Therefore,
[tex]\Delta{x}=\frac{1-0}{50}=\frac{1}{50}[/tex]
We need to divide the interval [0,1] into n = 50 sub-intervals of length [tex]\Delta{x}=\frac{1}{50}[/tex], with the following endpoints:
[tex]a=0, \frac{1}{50}, \frac{1}{25},...,\frac{24}{25}, \frac{49}{50}, 1=b[/tex]
Now, we just evaluate the function at these endpoints:
[tex]f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1[/tex]
[tex]2f\left(x_{1}\right)=2f\left(\frac{1}{50}\right)=2 e^{\frac{3}{25}}=2.25499370315875[/tex]
[tex]2f\left(x_{2}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281[/tex]
...
[tex]2f\left(x_{49}\right)=2f\left(\frac{49}{50}\right)=2 e^{\frac{147}{25}}=715.618483417705[/tex]
[tex]f\left(x_{50}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735[/tex]
Applying the trapezoid rule formula we get
[tex]\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{100}(1+2.25499370315875+2.54249830064281+...+715.618483417705+403.428793492735) \approx 67.1519320308594[/tex]
b. To approximate the integral [tex]\int_{0}^{1}e^{6 x}\ dx[/tex] using 2n with the Simpson's rule you must:
The Simpson's rule states that
[tex]\int_{a}^{b}f(x)dx\approx \\\frac{\Delta{x}}{3}\left(f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)\right)[/tex]
where [tex]\Delta{x}=\frac{b-a}{n}[/tex]
We have that a = 0, b = 1, n = 50
Therefore,
[tex]\Delta{x}=\frac{1-0}{50}=\frac{1}{50}[/tex]
We need to divide the interval [0,1] into n = 50 sub-intervals of length [tex]\Delta{x}=\frac{1}{50}[/tex], with the following endpoints:
[tex]a=0, \frac{1}{50}, \frac{1}{25},...,\frac{24}{25}, \frac{49}{50}, 1=b[/tex]
Now, we just evaluate the function at these endpoints:
[tex]f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1[/tex]
[tex]4f\left(x_{1}\right)=4f\left(\frac{1}{50}\right)=4 e^{\frac{3}{25}}=4.5099874063175[/tex]
[tex]2f\left(x_{2}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281[/tex]
...
[tex]4f\left(x_{49}\right)=4f\left(\frac{49}{50}\right)=4 e^{\frac{147}{25}}=1431.23696683541[/tex]
[tex]f\left(x_{50}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735[/tex]
Applying the Simpson's rule formula we get
[tex]\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{150}(1+4.5099874063175+2.54249830064281+...+1431.23696683541+403.428793492735) \approx 67.0715427161943[/tex]
c. If B is our estimate of some quantity having an actual value of A, then the absolute error is given by [tex]|A-B|[/tex]
The absolute error in the trapezoid rule is
The calculated value is
[tex]\int _0^1e^{6\:x}\:dx=\frac{e^6-1}{6} \approx 67.0714655821225[/tex]
and our estimate is 67.1519320308594
Thus, the absolute error is given by
[tex]|67.0714655821225-67.1519320308594|=0.08047[/tex]
The absolute error in the Simpson's rule is
[tex]|67.0714655821225-67.0715427161943|=0.00008[/tex]
A 20 year par value bond with semi-annual coupons at a nominal annual rate of 8% convertible semi-annually is purchased at a price of 1783.27. The bond can be called at par value X on any coupon date starting at the end of year 12 after the coupon is paid. The price guarantees a nominal annual rate of interest convertible semi-annually of at least 6%. Calculate X.
Answer:
3.216%
Step-by-step explanation:
This bond sells at a higher price or value, which means that its coupon is bogus of market interest rate. Therefore, the minimum yield rate that accounts for the possibility of the bond being called is calculated at the earliest possible call date. Let say exactly 15 years from the date of purchase, because that would be the most disadvantageous date for the bondholder for the call to occur.
The minimum semiannual yield:
j= i²/2
i² = 2j
which therefore satisfies the expression below for the worst possible case scenario yield:
1722.25 = 0.04*1100*[tex]a]_30[/tex]+[tex]\frac{1100}{(1+j)^30}[/tex]
Also, with the use of a financial calculator (making sure that the calculator is not in BGN mode)
1722.25 PV, -44 PMT, -1100 FV, 30 N, CPT 1/Y.
j can be found to be 1.608245%. The corresponding nominal annual rate compounded semiannually is (X) = i² = 2j =3.216%
Consider the following data collected in two recent surveys of whether voters in cities A and B favor a ballot proposition in the next election. City Sample Size In Favor A 615 463 B 585 403 Suppose you're going to find a confidence interval for the difference between the population proportions in the two cities. What's the standard error of the estimate of the difference between the two proportions?
Answer:
Standard error of the estimate of the difference between the two proportions=0.0259
Step-by-step explanation:
Given that the following data collected in two recent surveys of whether voters in cities A and B favor a ballot proposition in the next election.
City A B Total
Sample size 615 585 1200
Favour X 463 403 866
Proportion p 0.7528 0.6889 0.7217
Std error for difference
= [tex]\sqrt{p(1-p)(\frac{1}{n_1} }+ \frac{1}{n_2} \\[/tex]
p =0.7217
1-p = 0.2783
by substituting p and n1 = 615 and n2 = 585 we get
Std error = 0.0259
Standard error of the estimate of the difference between the two proportions=0.0259
drug that is used for treating cancer has potentially dangerous side effects if it is taken in doses that are larger than the required dosage for the treatment. The pharmaceutical company that manufactures the drug must be certain that the standard deviation of the drug content in the tablet is not more than 0.1 mg. Twenty-five tablets are randomly selected and the amount of drug in each tablet is measured. The sample has a mean of 20 mg and a variance of 0.02 mg. The hypotheses for the test are H0: ?2 ? 0.01 vs Ha: ?2 > 0.01.
Step 1 of 2:
Calculate the test statistic. Round your answer to two decimal places.
Answer:
[tex] t=(25-1) [\frac{0.141}{0.1}]^2 =47.71[/tex]
Step-by-step explanation:
Data given
[tex]\bar X=20[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
[tex]s^2=0.02[/tex] represent the sample variance
[tex]s=0.141[/tex] represent the sample deviation
n=25 represent the sample size
State the null and alternative hypothesis
On this case we want to check if the population standard deviation is more than 0.01, so the system of hypothesis are:
H0: [tex]\sigma \leq 0.1[/tex]
H1: [tex]\sigma >0.1[/tex]
In order to check the hypothesis we need to calculate the statistic given by the following formula:
[tex] t=(n-1) [\frac{s}{\sigma_o}]^2 [/tex]
This statistic have a Chi Square distribution distribution with n-1=25-1=24 degrees of freedom.
What is the value of your test statistic?
Now we have everything to replace into the formula for the statistic and we got:
[tex] t=(25-1) [\frac{0.141}{0.1}]^2 =47.71[/tex]
What is the critical value for the test statistic at an α = 0.05 significance level?
Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with 24 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.
We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.95,24)". And our critical value would be [tex]\chi^2 =36.415[/tex]
Since our calculated value is higher than the critical value we reject the null hypothesis at 5% of significance.
The variance hypothesis test for a cancer treatment drug with a sample mean of 20 mg and sample variance of 0.02 mg results in a chi-square test statistic of 48. This test statistic will be used to determine if the drug's variance exceeds the acceptable limit.
Explanation:The question at hand is concerning a hypothesis test of the variance in dosage of a cancer treatment drug. The null hypothesis (H0) claims that the standard deviation of the drug content is not more than 0.1 mg, which corresponds to a variance of 0.01 mg² since variance = standard deviation². The alternative hypothesis (Ha) is that the variance is greater than 0.01 mg². Given the sample variance as 0.02 mg² and a sample size of 25, the test statistic for the chi-square test can be calculated using the formula:
Test statistic (chi-square) = (n - 1)*sample variance / hypothesized variance
Test statistic = (25 - 1) * 0.02 / 0.01 = 24 * 2 = 48
The calculated test statistic is 48. Since the sample variance is greater than the hypothesized variance, we have a test statistic that would fall in the rejection region based on the selected significance level in a Chi-square distribution, suggesting that the drug dosage may indeed have greater variability than the company's standard.
c. Assume that the manufacturer's claim is true, what is the probability of such a tire lasting more than 60,000 miles?
Answer:
The probability of such a tire lasting more than 60,000 miles is 0.0228, for the complete question provided in explanation.
Step-by-step explanation:
Q. This is the question:The lifetime of a certain type of car tire are normally distributed. The mean lifetime of a car tire is 50,000 miles with a standard deviation of 5,000 miles. Assume that the manufacturer's claim is true, what is the probability of such a tire lasting more than 60,000 miles?
Answer:This is the question of normal distribution:
First w calculate the value of Z corresponding to X = 60,000 miles
We, have; Mean = μ = 50,000 miles, and Standard Deviation = σ = 5,000 miles
Now, for Z, we know that:
Z = (x-μ)/σ
Z = (60,000 - 50,000)/5,000
Z = 2
Now, we have standard tables, for normal distribution in terms of values of Z. One is attached in this answer.
P(X > 60,000) = P(Z > 2) = 1 - P(Z = 2)
P(X > 60,000) = 1 - 0.9772
P(X > 60,000) = 0.0228
There are 4 suits (heart, diamond, clover, and spade) in a 52-card deck, and each suit has 13 cards. Suppose your experiment is to draw one card from a deck and observe what suit it is. Express the probability in fraction format. (Show all work. Just the answer, without supporting work, will receive no credit.)
Answer:
The probability of drawing a heart or diamond is 1/2 or 0.5
The probability that the card is not a spade is 3/4 or 0.75
Step-by-step explanation:
Consider the provided information.
Part (a) Find the probability of drawing a heart or diamond.
There are 13 cards of heart and 13 cards of diamond.
We need to find the probability of drawing a heart or diamond.
[tex]P(\text{Heart or Diamond})=P(\text{Heart card Drawn})+P(\text{Diamond card Drawn})[/tex]
[tex]P(\text{Heart or Diamond})=\frac{13}{52}+\frac{13}{52}[/tex]
[tex]P(\text{Heart or Diamond})=\frac{26}{52}=\frac{1}{2}=0.5[/tex]
Hence, the probability of drawing a heart or diamond is 1/2 or 0.5
(b) Find the probability that the card is not a spade.
Out of 52 cards 13 are spade,
That means 52 - 13 = 39 cards are not a spade.
[tex]P(\text{Not spade})=\frac{39}{52}=\frac{3}{4}=0.75[/tex]
Hence, the probability that the card is not a spade is 3/4 or 0.75
Evaluate the expression for the given values of the variables.
Evaluate 9p − 8q for p = 4 and q = −8.
The expression is equal to .
The expression 9p − 8q for p = 4 and q = −8 is equal to 100
Solution:Given that we have to evaluate expression for the given values of the variables
Expression is:
⇒ 9p − 8q
Given that p = 4 and q = -8
Let us substitute the given values of p = 4 and q = -8 in given expression and evaluate it
⇒ 9p − 8q = 9(4) - 8(-8)
⇒ 9p - 8q = 9(4) + 8(8)
Upon multiplying the terms we get,
⇒ 9p - 8q = 36 + 64 = 100
Thus the expression is equal to 100
Write an equation in slope-intercept form of the line having the given slope and y-intercept. m:-4/6, (0,-4)
Answer:
y = -4/6x - 4
Step-by-step explanation:
y = m(x - x₁) + y₁
You're given m=-4/6 and (0,-4) ←x₁=0, y₁=-4
so just plug it into the point-slope equation.
y = (-4/6)(x - (0)) + (-4)
y = (-4/6)(x) + (-4)
y = -4/6x - 4
Answer:y = -4x/6 - 4
Step-by-step explanation:
The equation of a straight line can be represented in the slope intercept form as
y = mx + c
Where
m = slope = (change in the value of y in the y axis) / (change in the value of x in the x axis)
The slope,m of the given line is -4/6
To determine the intercept, we would substitute m = -4/6, x = 0 and y = -4 into y = mx + c. It becomes
- 4 = -4/6 × 0 + c = 0 + c
c = - 4
The equation becomes
y = -4x/6 - 4