An article reported that for a sample of 58 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO2 level (ppm) was 654.16, and the sample standard deviation was 165.4.

(a) Calculate and interpret a 95% (two-sided) confidence interval for true average CO2 level in the population of all homes from which the sample was selected. (Round your answers to two decimal places.) , ppm Interpret the resulting interval. We are 95% confident that the true population mean lies below this interval. We are 95% confident that this interval does not contain the true population mean. We are 95% confident that this interval contains the true population mean. We are 95% confident that the true population mean lies above this interval.
(b) Suppose the investigators had made a rough guess of 184 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 47 ppm for a confidence level of 95%?

Answer:

Step-by-step explanation:

Suppose we a point [tex]P(x,y,z)[/tex] such that its distance from either the point [tex]A(3,4,-5)[/tex] or [tex]B(-2,1,4)[/tex] is the same.

Using this information we can formula:

distance AP = distance BP

first, let's find the distance from AP, using the distance formula.

[tex]r = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}[/tex]

[tex]AP = \sqrt{(3 - x_2)^2 + (4 - y_2)^2 + (-5 - z_2)^2}[/tex]

similarly, we can find the distance BP

[tex]BP = \sqrt{(-2 - x_2)^2 + (1 - y_2)^2 + (4 - z_2)^2}[/tex]

since both distances are exactly the same we can equate them

[tex]AP = BP[/tex]

[tex]\sqrt{(3 - x_2)^2 + (4 - y_2)^2 + (-5 - z_2)^2} = \sqrt{(-2 - x_2)^2 + (1 - y_2)^2 + (4 - z_2)^2}[/tex]

we can simplify it a bit squaring both sides, and getting rid of the subscripts.

[tex](3 - x)^2 + (4 - y)^2 + (-5 - z)^2 = (-2 - x)^2 + (1 - y)^2 + (4 - z)^2[/tex]

what we have done here is formulated an equation which consists of any point P that will have the same distance from (3,4,-5) and (-2,1,4).

To put it more concretely,

This is the equation of the the plane from that consists of all points (P) from which the distance from both (3,4,-5) and (-2,1,4) are equal.

Answer:

First question:

Top 6%: 4.87 ounces

Bottom 6%: 4.75 ounces

Second question:

Top 7%: Score of 649.4.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

For the first problem, we have that:

[tex]\mu = 4.81, \sigma = 0.04[/tex]

Top 6%

The value of X when Z has a pvalue of 0.94. This is [tex]Z = 1.555[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.555 = \frac{X - 4.81}{0.04}[/tex]

[tex]X - 4.81 = 1.555*0.04[/tex]

[tex]X = 4.8722[/tex]

Bottom 6%

The value of X when Z has a pvalue of 0.06. This is [tex]Z = 1.555[/tex]

For the second problem, we have that:

[tex]\mu = 496, \sigma = 109[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-1.555 = \frac{X - 4.81}{0.04}[/tex]

[tex]X - 4.81 = -1.555*0.04[/tex]

[tex]X = 4.7477[/tex]

Top 7%

The value of X when Z has a pvalue of 0.93. This is [tex]Z = 1.475[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.475 = \frac{X - 496}{104}[/tex]

[tex]X - 496 = 104*1.475[/tex]

[tex]X = 649.4[/tex]

Answer: (0.089, 0.125)

Step-by-step explanation:

Confidence interval for population proportion is given by :-

[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

, where n= sample size.

[tex]\hat{p}[/tex] = Sample proportion.

z*= Critical z-value.

Let p be the population proportion of people the company contacts who may buy something.

As per given , sample size : n= 1140

Number of recipients ordered = 122

Then, [tex]\hat{p}=\dfrac{122}{1140}\approx0.107[/tex]

Critical value for 95% confidence interval = z*= 1.96 (By z-table)

So , the 95% confidence interval for the percentage of people the company contacts who may buy something:

[tex]0.107\pm (1.96)\sqrt{\dfrac{0.107(1-0.107)}{1140}}[/tex]

[tex]=0.107\pm (1.96)\sqrt{0.000083817}[/tex]

[tex]=0.107\pm (1.96)(0.00915516)[/tex]

[tex]=0.107\pm 0.018[/tex]

[tex]=(0.107-0.018,\ 0.107+0.018)=(0.089,\ 0.125)[/tex]

Hence, the 95% confidence interval for the percentage of people the company contacts who may buy something = (0.089, 0.125)

Answer:

[tex] \frac{1}{X} \sim log N(-\mu , \sigma^2)[/tex]

Step-by-step explanation:

For this case we know that the distribution for the random variable is given by:

[tex]X \sim logN(\mu ,\sigma^2)[/tex]

The density function for the log normal random variable is given by:

[tex] f)x,\sigma) = \frac{1}{x \sigma \sqrt{2\pi}}e^{- \frac{ln x^2}{2\sigma^2}}[/tex]

And we want to find the distribution for the random variable [tex]\frac{1}{X}[/tex]

In order to find this distribution we can use the cumulative distribution function like this:

Let [tex] Y= \frac{1}{X}[/tex], if we solve for X from this transformation we got:

[tex] X= \frac{1}{Y}[/tex]

And then we have this:

[tex] F_Y (y) = P(Y \leq y) = P(X \leq \frac{1}{y}) = F_X (\frac{1}{y})[/tex]

And we can find the density function as the derivate of the distribution function like this:

[tex] f_Y (y) = F'_Y (y) = -\frac{1}{y^2} f_Y(\frac{1}{y})[/tex]

[tex] f_Y (y)= -\frac{1}{y^2} \frac{1}{\frac{1}{y} \sigma \sqrt{2\pi}} e^{- \frac{ln(\frac{1}{y})^2}{2\sigma^2}}[/tex]

But we see that we don't have an specified form for the distribution

If we assume that X follows a normal distribution [tex] X\sim N (\mu_z,\sigma^2_z)[/tex] and we use the transformation [tex]X=e^Y[/tex] we see that X follows a log normal distribution. And we see that:

[tex]\frac{1}{X}= \frac{1}{e^Y}=e^{-Y}[/tex]

And if we find the distribution of [tex]e^{-y}[/tex] we got this:

[tex] f_Y (y) = F'_Y (y) = -e^{-y} f_Y(e^{-y})[/tex]

[tex] f_Y (y)= -e^{-y} \frac{1}{e^{-y} \sigma \sqrt{2\pi}} e^{- \frac{ln(e^{-y})^2}{2\sigma^2}}[/tex]

[tex] f_Y (y) = -\frac{1}{\sigma \sqrt{2\pi}}e^{\frac{y^2}{2\sigma^2}}[/tex]

And then we see that [tex]Y= \frac{1}{X} \sim log N(-\mu , \sigma^2)[/tex]

Answer:

Option b) significantly different from 24

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 24

Sample mean, [tex]\bar{x}[/tex] = 25

Sample size, n = 25

Alpha, α = 0.05

Population standard deviation, σ = 2

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 24\text{ years}\\H_A: \mu \neq 24\text{ years}[/tex]

We use Two-tailed z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{25 - 24}{\frac{2}{\sqrt{16}} } = 2[/tex]

Now, we calculate the p-value from the standard normal table.

P-value = 0.0455

Since the p-value is less than the significance level, we fail to accept the null hypothesis and accept the alternate hypothesis.

Thus, we conclude that mean age is significantly different from 24.

At 5% level of significance, it can be concluded that the mean age is B. Significantly different from 24.

From the information given, the university has had an average age of 25 years. The critical value at 5% level with the degree of freedom is 1.753 while the test statistic will be:

= (25 - 24)/2 × ✓26

= 1/2 × 4 = 2

Since the test statistic is greater than 1.753, one should reject the null hypothesis. Therefore, it shows that that the mean age is significantly different from 24.

Learn more about significance level on:

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Answer:

Null hypothesis:[tex]p\leq 0.5[/tex]  

Alternative hypothesis:[tex]p > 0.5[/tex]  

[tex]z=\frac{0.765 -0.5}{\sqrt{\frac{0.5(1-0.5)}{51}}}=3.785[/tex]  

[tex]p_v =P(z>3.785)=7.68x10^{-5}[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of boys is significantly higher than 0.5.  

Step-by-step explanation:

1) Data given and notation

n=51 represent the random sample taken

X=39 represent the number of boys

[tex]\hat p=\frac{39}{51}=0.765[/tex] estimated proportion of boys

[tex]p_o=0.5[/tex] is the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that with this method, the probability of a baby being a boy is greater than 0.5.:  

Null hypothesis:[tex]p\leq 0.5[/tex]  

Alternative hypothesis:[tex]p > 0.5[/tex]  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.765 -0.5}{\sqrt{\frac{0.5(1-0.5)}{51}}}=3.785[/tex]  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

[tex]p_v =P(z>3.785)=7.68x10^{-5}[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of boys is significantly higher than 0.5.  

Answer:Jarvis's class average is 75

Step-by-step explanation:

The total possible average score for the math course is 100

a) If the teacher rates homework at 10%, it means that the total possible score for homework

is 10/100 × 100 = 10

If his homework average is 93, then his score would be

(93×10)/100 = 9.3

b) If the teacher rates quizzes at 30%, it means that the total possible score for quizzes

is 30/100 × 100 = 30

If his quiz average is 82, then his score would be

(82×30)/100 = 24.6

c) If the teacher rates test at 40%, it means that the total possible score for quizzes

is 40/100 × 100 = 40

If his quiz average is 72, then his score would be

(72×40)/100 = 28.8

d) If the teacher rates final exam at 20%, it means that the total possible score for quizzes

is 20/100 × 100 = 20

If his final exam is 60, then his score would be

(60×20)/100 = 12

Jarvis's class average would be

9.3 + 24.6 + 28.8 + 12 = 74.7

Approximately 75

Answer: [tex](98.04,\ 98.36)[/tex]

Step-by-step explanation:

Given : Sample size of healthy adults: n= 106

Degree of freedom = df =n-1 = 105

Sample mean body temperature  : [tex]\overline{x}=98.2[/tex]

Sample standard deviation : [tex]s= 0.62[/tex]

Significance level ; [tex]\alpha= 1-0.99=0.01[/tex]

∵ population standard deviation is unknown , so we use t- critical value.

Confidence interval for the population mean :

[tex]\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}[/tex]

Using t-distribution table , we have

Critical value for df = 105 and [tex]\alpha=0.01[/tex]= [tex]t_{\alpha/2, df}=t_{0.005 , 105}=2.623[/tex]

A 99% confidence interval estimate of the mean body temperature of all healthy humans will be :

[tex]98.2\pm (2.623)\dfrac{0.62}{\sqrt{106}}[/tex]

[tex]98.2\pm (2.623)(0.0602197234662)[/tex]

[tex]98.2\pm (0.157956334652)[/tex]

[tex]\approx98.2\pm 0.16[/tex]

[tex](98.2-0.16,\ 98.2+0.16)[/tex]

[tex](98.04,\ 98.36)[/tex]

Hence, a 99% confidence interval estimate of the mean body temperature of all healthy humans = [tex](98.04,\ 98.36)[/tex]

Answer:

There is no sufficient evidence to support the claim that wedding cost is less than $30000.

Step-by-step explanation:

Values (x) ∑(Xi-X)^2

----------------------------------

29.1                    0.1702

28.5                  1.0252

28.8                  0.5077

29.4                   0.0127

29.8                  0.0827

29.8                  0.0827

30.1                   0.3452

30.6                   1.1827

----------------------------------------

236.1                 3.4088

Mean = 236.1 / 8 = 29.51

[tex]S_{x}=\sqrt{3.4088/(8-1)}=0.6978[/tex]

Statement of the null hypothesis:

H0: u ≥ 30 the mean wedding cost is not less than $30,000

H1: u < 30 the mean wedding cost is less than $30,000

Test Statistic:

[tex]t=\frac{X-u}{S/\sqrt{n}}=\frac{29.51-30}{0.6978/\sqrt{8}}= \frac{-0.49}{0.2467}=-1.9861[/tex]

Test criteria:

SIgnificance level = 0.05

Degrees of freedom = df = n - 1 = 8 - 1 = 7

Reject null hypothesis (H0) if

[tex]t<-t_{0.05,n-1}\\ t<-t_{0.05,8-1}\\ t<-t_{0.05,7}[/tex]

Finding in the t distribution table α=0.05 with df=7, we have

[tex]t_{0.05,7}=2.365[/tex]

[tex]t>-t_{0.05,7}[/tex] = -1.9861 > -2.365

Result: Fail to reject null hypothesis

Conclusion: Do no reject the null hypothesis

u ≥ 30 the mean wedding cost is not less than $30,000

There is no sufficient evidence to support the claim that wedding cost is less than $30000.

Hope this helps!

Answer: 2172

Step-by-step explanation:

Formula to find the sample size n , if the prior estimate of the population proportion(p) is known:

[tex]n= p(1-p)(\dfrac{z^*}{E})^2[/tex] , where E=  margin of error and z* = Critical z-value.

Let p be the population proportion of adults have consulted fortune tellers.

As per given , we have

p= 0.20

E= 0.02

From z-table , the z-value corresponding to 98% confidence interval = z*=2.33

Then, the required sample size will be :

[tex]n= 0.20(1-0.20)(\dfrac{2.33}{0.02})^2[/tex]

[tex]n= 0.20(0.80)(116.5)^2[/tex]

[tex]n= 2171.56\approx2172[/tex]

Hence, the required sample size = 2172

Answer:

P-value =  0.4846

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 4.73

Sample mean, [tex]\bar{x}[/tex] = 4.35

Sample size, n = 51

Alpha, α = 0.05

Sample standard deviation, s = 3.88

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 4.73\\H_A: \mu \neq 4.73[/tex]

We use Two-tailed z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{4.35 - 4.73}{\frac{3.88}{\sqrt{51}} } = -0.699[/tex]

Now, [tex]z_{critical} \text{ at 0.05 level of significance } = \pm 1.96[/tex]

We can calculate the p-value with the help of standard normal table.

P-value =  0.4846

Since the p-value is higher than the significance level, we fail to reject the null hypothesis and accept it.

We conclude that this college has same drinking habit as the college students in general.

Answer:

An independent samples t-test

Step-by-step explanation:

We have two different groups and we want to test if the scores are equal or not , so the best appropiate mthod would be an independent t test. Here we show the steps to conduct this test.

Data given and notation  

[tex]\bar X_{1}[/tex] represent the mean for group men  

[tex]\bar X_{2}[/tex] represent the mean for group women  

Assuming these values for the remaining data:

[tex]s_{1}[/tex] represent the sample standard deviation for men

[tex]s_{2}[/tex] represent the sample standard deviation for women

[tex]n_{1}[/tex] sample size for the group men  

[tex]n_{2}[/tex] sample size for the group women  

t would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value  

Concepts and formulas to use  

Suppose that we need to conduct a hypothesis in order to check if the mean are equal or not, the system of hypothesis would be:  

H0:[tex]\mu_{1} = \mu_{2}[/tex]  

H1:[tex]\mu_{1} \neq \mu_{2}[/tex]  

For this case is better apply a t test to compare means since we don't know the population deviations, and the statistic is given by:  

[tex]z=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}[/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.  

Calculate the statistic  

We just need to replace in formula (1) and find the calculated value.    

Find the critical value

In order to find the critical value we need to take in count that we are conducting a two tailed test, and we need a significance level provided in order to find the critical region

Statistical decision

If our calculates value [tex]t_{calculated}>t_{critical}[/tex] or [tex]t_{calculated}<t_{critical}[/tex] we reject the null hypothesis. In other case we FAIL to reject the null hypothesis.

Answer:

Confidence interval for the intercept:

[tex]37.67-2.1123(5.69) \leq \beta_0 \leq 37.67+2.1123(5.69)\\25.651 \leq \beta_0 \leq 49.6889[/tex]

Confidence interval for the slope:

[tex]33.18-2.1123(7.94) \leq \beta_1 \leq 33.18+2.1123(7.94)\\16.4083 \leq \beta_1 \leq 49.9517[/tex]

Step-by-step explanation:

We start defining our equation's terms, starting from the linear regression model [tex]\hat{y} =\hat{\beta}_{0} + \hat{\beta}_{1}x[/tex]

In this model [tex]{\beta}_{0} [/tex] is the intercept estimator and [tex]{\beta}_{1} [/tex] is the slope estimator.

in the equation y = 37.67 + 33.18x]

[tex]{\beta}_{0} = 37.67 [/tex] and [tex]{\beta}_{1} = 33.81[/tex]

Then we have the standard errors (se) for each estimator:

[tex]se(\hat{\beta_{0}}}) =  5.69[/tex] and [tex]se(\hat{\beta_{1}}}) =  7.94[/tex]

The sample number is 49 species (here we assume that all the individuals of a the same species are summarised with a central tendency measure, i.e. mean, median or mode, if each species contained more than one individual).

The formula for the 96% confidence interval for the intercept [tex]{\beta}_{0} [/tex] we have:

[tex]\hat{\beta_0} - t_{\alpha/2,n-2} se(\beta_0) \leq  \beta _0 \leq \hat{\beta_0} + t_{\alpha/2,n-2}[/tex]

Where  [tex]t_{\alpha/2,n-2}[/tex] represents the p-value in a t distribution when α=0.04 (so that we have a confidence interval of 96%, or 0.96), two-tailed, and n-2 degrees of freedom. In this example, n-2 = 47, and the t-value (47 degrees of freedom, 0.04 significance level, two tails) is ± 2.1123.

We input these values into our formula:

[tex]37.67-2.1123(5.69) \leq \beta_0 \leq 37.67+2.1123(5.69)\\25.651 \leq \beta_0 \leq 49.6889[/tex]

Similarly, the 96% confidence interval for the slope  [tex]{\beta}_{1} [/tex] is:

[tex]\hat{\beta_1} - t_{\alpha/2,n-2} se(\beta_1) \leq  \beta _1 \leq \hat{\beta_1} + t_{\alpha/2,n-2}[/tex]

Where [tex]t_{\alpha/2,n-2} =± 2.1123[/tex]

And into the formula:

[tex]33.18-2.1123(7.94) \leq \beta_1 \leq 33.18+2.1123(7.94)\\16.4083 \leq \beta_1 \leq 49.9517[/tex]

The confidence interval does not include 0, so there is enough evidence saying that there is enough correlation between the metatarsal-to-femur ratio and maximal sprint speed in kilometers per hour. This study shows that measuring the lengths of metatarsal 3 and femur in mammals is a reliable predictor of maximal speed for cursorial mammals.

The 96% confidence interval for the slope of the population regression line is calculated using the formula b ± t(α/2, df) * SE(b). Plugging in the given values (slope= 33.18, standard error= 7.94) along with t(0.02, 47) which is approximately ± 2.01, we get the interval (16.98, 49.38).

In statistics, the confidence interval for the slope of the population regression line can be estimated by using the following formula: b ± t(α/2, df) * SE(b), where b is the slope, t is the t-score, df represents the degrees of freedom, α is the significance level, and SE(b) is the standard error of the slope.

In this case, we are asked to construct a 96% confidence interval for the slope, so α = 1 - 0.96 = 0.04. Since we are dealing with a two-tailed test, we will distribute α across the two tails, giving us α/2 = 0.02 for each half. To calculate our t-score, we will use the t distribution table and look for the value that corresponds with df = n - 2 (since we have two parameters, the intercept and slope) and α/2 = 0.02. Given we have 49 species, the df = 49 - 2 = 47, the t-score is approximately ± 2.01.

With this information, we can substitute the values into the formula, where b = 33.18 and SE(b) = 7.94, and compute the confidence interval for the slope, giving us 33.18 ± 2.01 * 7.94 = (16.98, 49.38). Thus, we estimate with 96% confidence that the true population slope lies between 16.98 and 49.38.

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Answer:

[tex]p_v =P(t_{17}<-1.728)=0.051[/tex]    

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

The best option for this case would be:

We would switch to α = .01 for a more powerful test.

Step-by-step explanation:

Previous concepts and data given  

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".  

[tex]\bar X=37.8[/tex] represent the sample mean  

[tex]s=5.4[/tex] represent the sample standard deviation  

n=18 represent the sample selected  

[tex]\alpha=0.05[/tex] significance level  

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is less than 40, the system of hypothesis would be:    

Null hypothesis:[tex]\mu \geq 40[/tex]    

Alternative hypothesis:[tex]\mu < 40[/tex]    

If we analyze the size for the sample is < 30 and we don't know the population deviation so 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{37.8-40}{\frac{5.4}{\sqrt{18}}}=-1.728[/tex]      

P-value

First we need to calculate the degrees of freedom given by:

[tex]df=n-1=18-1=17[/tex]

Since is a left tailed test the p value would be:    

[tex]p_v =P(t_{17}<-1.728)=0.051[/tex]    

Conclusion    

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

The best option for this case would be:

We would switch to α = .01 for a more powerful test.

Since the p values is just a little higher than th significance level 0.051>0.05  but the values are two close. If we change the value of the significance by 0.01, we have that 0.051>0.01 and it's a more powerful test.

Final answer:

To find the variance of a binomial distribution with n = 800 and p = 0.87, use the formula Variance = n × p × (1-p). The variance in this case is approximately 90.64.

Explanation:

To find the variance of a binomial distribution, we use the formula:

Variance = n × p × (1-p)

Where n is the number of trials and p is the probability of success.

So, in this case, with n = 800 and p = 0.87, we can calculate the variance as:

Variance = 800 × 0.87 ×  (1-0.87)

Variance = 800 × 0.87 × 0.13

Variance = 90.64

Therefore, the variance of the binomial distribution is approximately 90.64.

Answer:

First

Step-by-step explanation:

Since, the probability that the first person pick the paper marked "you lose" is 1/200, which is smaller than the probability of who draws later.

Taking your turn to draw last is the most strategic choice to minimize the risk of being the unfortunate person to draw the "you lose" paper.

The optimal strategy in this scenario is to draw last.

Here's the reason:

When drawing first, you have a 1/200 chance of drawing the "you lose" paper initially.

However, if you draw last, you get to observe the outcomes of all the previous draws.

If none of the previous guests has drawn the "you lose" paper, then the odds of it being in the hat when it's your turn are 1/200.

In contrast, if someone before you draws the "you lose" paper, you won't have to draw at all since the game ends. This means that by drawing last, you have the best chance of avoiding the "you lose" paper if others before you didn't draw it.

Thus, taking your turn to draw last is the most strategic choice to minimize the risk of being the unfortunate person to draw the "you lose" paper.

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Answer:

a) Sample mean: 48.71

Sample median: 48.3

Sample variance: 5.41

Sample standard deviation: 2.37

Step-by-step explanation:

a) Sample mean:

[tex]\bar{X}=\frac{1}{N} \sum X_i=\frac{1}{80}*3896.9=48.71[/tex]

Sample median: M=48.3

Note: I order the data increasingly and take the value N / 2 = 40. In this way there are 39 values above and 39 values below the median.

Sample variance:

[tex]s^2=\frac{1}{N-1}\sum (X_i-\bar X)^2 =(\frac{1}{80-1})*427.74=5.41[/tex]

Sample standard deviation

[tex]s=\sqrt{s^2}=\sqrt{5.41}=2.37[/tex]

b)

Answer:

integers are closed under division

Step-by-step explanation:

Integers are not closed under division.Let us consider two integers i and j.

Then [tex]\[\frac{i}{j}\][/tex] need not necessarily be an integer. For example, 2 and 3 are integers but [tex]\[\frac{2}{3}\][/tex] is not an integer. So the first option , namely, integers are closed under division is not true.

On the other hand, the remaining three options given are correct.

Answer:

The 99% confidence interval would be given (0.004;0.436).

Answer:

[tex]P(600<\bar X<700)=0.89347[/tex]

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the total amounts spent by students vacationing for a week in Florida of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(650,120)[/tex]  

Where [tex]\mu=650[/tex] and [tex]\sigma=120[/tex]

And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:

[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]

On this case  [tex]\bar X \sim N(650,\frac{120}{\sqrt{15}})[/tex]

We are interested on this probability

[tex]P(600<\bar X<700)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

If we apply this formula to our probability we got this:

[tex]P(600<\bar X<700)=P(\frac{600-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{700-\mu}{\frac{\sigma}{\sqrt{n}}})[/tex]

[tex]=P(\frac{600-650}{\frac{120}{\sqrt{15}}}<Z<\frac{700-650}{\frac{120}{\sqrt{15}}})=P(-1.614<Z<1.614)[/tex]

And we can find this probability on this way:

[tex]P(-1.614<Z<1.614)=P(Z<1.614)-P(Z<-1.614)[/tex]

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

[tex]P(-1.614<Z<1.614)=P(Z<-1.614)-P(Z<-1.614)=0.94674-0.05326=0.89347[/tex]

To find the probability, calculate the z-scores for $600 and $700, and find the area under the normal curve between these two z-scores.

To find the probability that a simple random sample (SRS) of 15 students will spend an average of between $600 and $700, we need to calculate the z-scores for these two values.

First, we find the z-score for $600:

z = (600 - 650) / (120 / √15) = -1.5496

Next, we find the z-score for $700:

z = (700 - 650) / (120 / √15) = 1.5496

We then use a standard normal table or calculator to find the area under the normal curve between these two z-scores. The probability is the difference between these two areas.

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Answer:

Line segment RS

Step-by-step explanation:

Hence the correct answer is Line segment RS.

Answer:

a) [tex]t_{crit}=1.34[/tex]

b) [tex]t_{crit}=-1.33[/tex]

c) [tex]t_{crit}=\pm 2.11[/tex]

Step-by-step explanation:

Part a

[tex]\alpha=0.1[/tex] represent the significance level

df =15

Since is a right tailed test the critical value is given by:

[tex]t_{crit}=1.34[/tex]

And we can use the following excel code to find it: "=T.INV(0.9,15)"

Part b

[tex]\alpha=0.1[/tex] represent the significance level

n=20 represent the sample size

First we need to find the degrees of freedom given by:

[tex]df=n-1=20-1=19[/tex]

Since is a left tailed test the critical value is given by:

[tex]t_{crit}=-1.33[/tex]

And we can use the following excel code to find it: "=T.INV(0.1,19)"

Part c

[tex]\alpha=0.05[/tex] represent the significance level

n=18 represent the sample size

First we need to find the degrees of freedom given by:

[tex]df=n-1=18-1=17[/tex]

The value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]

Since is a two tailed tailed we have two critical values is given by:

[tex]t_{crit}=\pm 2.11[/tex]

And we can use the following excel code to find it: "=T.INV(0.025,17)"

Answer:

C.ModifyingAbove p with caret p=0.224. The symbol ModifyingAbove p with caret p is used to represent a sample proportion.

Step-by-step explanation:

Data given and notation

X= 459 represent the number of people that answer yes

1241 people answer no and 352 had no opinion

n= 459+1241+352=2052 represent the sampplpe size selected

Solution to the problem

What is the sample proportion of yes​ responses, and what notation is used to represent​ it?

If we want to find the sample proportion of yes we need to apply the following formula:

[tex]\hat p =\frac{X}{n}[/tex]

Where X represent the number of people with the characteristic desired in the sample and n the sample size selected.

For this case we have:

[tex]\hat p =\frac{X}{n}=\frac{459}{2052}=0.224[/tex]

Let's analyze the possible options given:

A.ModifyingAbove p with caret p=0.270. The symbol ModifyingAbove p with caret p is used to represent a sample proportion.

False the sample proportion is 0.224 not 0.270

B.p=0.224. The symbol p is used to represent a sample proportion.

The sample proportion is 0.224 not the population proportion for this case this statement is False.

C.ModifyingAbove p with caret p=0.224. The symbol ModifyingAbove p with caret p is used to represent a sample proportion.

Correct the [tex]\hat p= 0.224[/tex] represent th sample proportion and it's the correct way to express it.

D.p=0.270. The symbol p is used to represent a sample proportion.

False the sample proportion is 0.224 not 0.270

To find the sample proportion of yes responses, divide the number of yes responses by the total responses. The sample proportion is 0.224, represented as p'. The correct answer is option C.

To determine the sample proportion of yes responses in the Harris poll, we need to use the formula for sample proportion:

p' = x/n, where x represents the number of successes (yes responses) and n represents the sample size (total responses).

Step-by-Step:

The sample proportion of yes responses is 0.224. In statistical notation, this is represented as p'.

The correct answer is: C. p' = 0.224. The symbol p' is used to represent a sample proportion.

Answer:

The probability that a student earns a grade of A is 1/7.

Step-by-step explanation:

Let E be an event and S be the sample space. The probability of E, denoted by P(E) could be computed as:

P(E) = n(E) / n(S)

As the total number of students = n(S) = 35

Students getting the grade A = n(E) = 5

So, the probability that a student earns a grade of A:

                      P(E) = n(E) / n(S)

                              = 5/35

                              = 1/7

Hence, the probability that a student earns a grade of A is 1/7.

Keywords: probability, sample space, event

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Answer:

.14

Step-by-step explanation:

Answer:

(B) The citizens of the city, because they lose help they could have used to buy a home.

Step-by-step explanation:

Nul and alternative hypotheses are:

Type II error occurs when one fails to reject null hypothesis when the null hypothesis was wrong.

In this case Type II error happens when the conclusion is the rate of home ownership is not increasing after tax cut, where actually it is.

With this conclusion city council does not continue tax cut, and citizens of the city is harmed because they lose help they could have used to buy a home.

Final answer:

The citizens of the city would be harmed by a Type II error as they would miss out on the help to buy homes.

Explanation:

Who would be harmed by a Type II error?

(B) The citizens of the city, because they lose help they could have used to buy a home.

A Type II error in this context would harm the citizens of the city as they would miss out on the intended help in buying homes due to the failure to detect an increase in the rate of home ownership.

Answer:

-uniform probability distribution

Step-by-step explanation:

In uniform probability distributions, the likelihood of each possible outcome happening or not is the same. This property means that, for any given trial, the probability that an event will be successful does not change. Take the probability of rolling a 5 on a die for instance, no matter how many trials are performed, there is always a 1 in 6 probability for each trial.

A probability distribution showing the probability of success does not change from trial to trial, is termed as a uniform probability distribution.

We have to determine, a probability distribution showing the probability of success does not change from trial to trial, is termed as;

A continuous probability distribution is called the uniform distribution and it is related to the events that are equally possible to occur. It is defined by two different parameters, x, and y,

Where x = the minimum value and y = the maximum value. It is generally represented by u(x, y).

The case of a discrete binomial probability distribution. The Bernoulli trials are identical but independent of each other. Before computing the failures (“r”), the total count of success that occurs first is called the Negative Binomial Probability Distribution.

Probability Distributions give up the possible outcome of any random event. It is also identified on the grounds of underlying sample space as a set of possible outcomes of any random experiment.

The normal distribution is a probability distribution that outlines how the values of a variable are distributed. Most of the observations cluster around the central peak and probabilities for values far away from the mean fall off equally in both directions in a normal distribution. Extreme values in both the tails of the distribution are uniformly unexpected.

Hence, A probability distribution showing the probability of success does not change from trial to trial, is termed as a uniform probability distribution.

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Answer:

[tex]\chi^2 = \frac{(141-141)^2}{141}+\frac{(291-282)^2}{282}+\frac{(132-141)^2}{141}=0.862[/tex]

[tex]p_v = P(\chi^2_{2} >0.861)=0.6502[/tex]

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(0.861,2,TRUE)"

Since the p value is higher than the significance level [tex]0.6502>0.05[/tex] we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences for the proportions assumed.

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

Assume the following dataset:

White = 141, Pink = 291, Red= 132

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference with the proportions assumed [tex]p_{white}=1/4, p_{pink}=1/2, p_{red}=1/4[/tex]

H1: There is difference with the proportions assumed [tex]p_{white}=1/4, p_{pink}=1/2, p_{red}=1/4[/tex]

The level of significance assumed for this case is [tex]\alpha=0.05[/tex]

The statistic to check the hypothesis is given by:

[tex]\chi^2 = \sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]

The table given represent the observed values, we just need to calculate the expected values with the following formula [tex]E_i = p_i *Total[/tex]

And the calculations are given by:

[tex]E_{White} =\fra{1}{4} * 564=141[/tex]

[tex]E_{Pink} =\frac{1}{2} *564=282[/tex]

[tex]E_{Red} =\frac{1}{4}*564=141[/tex]

And now we can calculate the statistic:

[tex]\chi^2 = \frac{(141-141)^2}{141}+\frac{(291-282)^2}{282}+\frac{(132-141)^2}{141}=0.862[/tex]

Now we can calculate the degrees of freedom for the statistic given by:

[tex]df=categories-1=3-1=2[/tex]

And we can calculate the p value given by:

[tex]p_v = P(\chi^2_{2} >0.861)=0.6502[/tex]

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(0.861,2,TRUE)"

Since the p value is higher than the significance level [tex]0.6502>0.05[/tex] we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences for the proportions assumed.

Answer:

[tex]P(\bar X < 5000)=P(Z<4 (0.2533))=P(Z<1.0132)=0.845[/tex]

Step-by-step explanation:

1) Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable that represent the mass of minerals of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(\mu,\sigma)[/tex]  

Where [tex]\mu=?[/tex] and [tex]\sigma=?[/tex]

From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

2) Solution to the problem

For this case we know this condition given :

[tex]P(X<5000)=0.6[/tex]

We can use the Z score given by this formula:

[tex]Z=\frac{X-\mu}{\sigma}[/tex]

And using this formula we got:

[tex]P(Z<\frac{5000-\mu}{\sigma})=0.6[/tex]

And we can find a value on the normal standard distribution that accumulates 0.6 of the are aon the left and 0.4 of the area on the right, on this case the value is Z=0.2533. And we can use the following excel code to find it :"=NORM.INV(0.6,0,1)"

So then we can do this:

[tex]0.2533=\frac{5000-\mu}{\sigma}[/tex]  (1)

By the other hand when we find the z score for the sample mean we have this:

[tex]Z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

And we want to find this probability:

[tex]P(\bar X < 5000)[/tex]

And if we use the z score formula we got:

[tex]P(Z< \frac{5000 -\mu}{\frac{\sigma}{\sqrt{16}}})=P(Z<\sqrt{16} \frac{5000-\mu}{\sigma})[/tex]  (2)

And replacing condition (1) into equation (2) we got:

[tex]P(Z<4 (0.2533))=P(Z<1.0132)=0.845[/tex]

And we can use the following excel code to find it: "=NORM.DIST(1.0132,0,1,TRUE)"

Answer:

The boys need to complete the run in 385.9 seconds or less in order to earn a certificate of recognition from the fitness association.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 450

Standard Deviation, σ = 50

We are given that the distribution of time for fitness test is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

We have to find the value of x such that the probability is 0.10

P(X<x) = 0.10

[tex]P( X < x) = P( z < \displaystyle\frac{x - 450}{50})= 0.10[/tex]

Calculation the value from standard normal z table, we have,  

[tex]P(z\leq -1.282) = 0.10[/tex]

[tex]\displaystyle\frac{x - 450}{50} = -1.282\\\\x = 385.9[/tex]

Hence, boys need to complete the run in 385.9 seconds or less in order to earn a certificate of recognition from the fitness association.

The boys need to beat a time of 514 seconds, which corresponds to the 90th percentile of the normal distribution with a mean of 450 seconds and a standard deviation of 50 seconds, to be in the fastest 10% and earn a certificate of recognition.

To determine the time the boys need to beat to be in the fastest 10% for the mile run, we need to find the corresponding z-score for the 90th percentile (since 100% - 10% = 90%) in a standard normal distribution. The z-score that correlates with the 90th percentile is approximately 1.28. We can then use the z-score formula:

Z = (X - μ) / σ

Where X is the time to beat, μ is the mean, and σ is the standard deviation. Substituting the values we know (μ = 450 seconds, σ = 50 seconds, and Z = 1.28):

1.28 = (X - 450) / 50

Multiplying both sides by 50 gives us:

64 = X - 450

Adding 450 to both sides gives:

X = 514 seconds

Therefore, boys need to beat a time of 514 seconds (or 8 minutes and 34 seconds) to earn a certificate of recognition.

Answer:

We accept the null hypothesis

Step-by-step explanation:

Given

The valve was tested on 120 engines

Mean pressure = 5 lbs/square inch.

Variance = 1

Standard Deviation, σ = √1 = 1

The valve was designed to produce a mean pressure of 5.1 lbs/square inch

So, μ = 5.1

Null hypothesis: H₀ : μ = 5.1

Alternative Hypothesis: H₁ : μ≠ 5.1

Since n > 30 and population standard deviation is given

So, We will use z test

Formula : z = (x - μ)/( σ /√n) ---_-_--- Substitute the values

z = [tex]\frac{5 - 5.1}{1/\sqrt{120} }[/tex]

z = −1.1

The p - value of -1.1 (refer to z table) is 0.13576

Since it is a two tailed test So, p = 2(1-  0.13576) = 1.7285

α = 0.02

p value > α

So, we accept the null hypothesis

Hence There is no sufficient evidence at the 0.02 level that the valve does not performs below to the specifications