In an advertisement, a pizza shop claims its mean delivery time is 30 minutes. A consumer group believes that the mean delivery time is greater than the pizza shop claims. A random sample of 41 delivery times has a mean of 31.5 minutes with standard deviation of 3.5 minutes. Does this indicate at a 5% significance level that the mean delivery time is longer than what the pizza shop claims?

15x+10y≤270 and x+y≥20 are the constraints for this situation.

Step-by-step explanation:

Given,

Cost of long sleeved shirt = $15

Cost of short sleeved shirt = $10

Amount to spend = $270

Shirts to order = 20

Let,

Long sleeved shirts = x

Short sleeved shirts = y

At most means he can not spend more than $270, therefore,

15x+10y≤270

At least 20 means, he needs minimum 20 or more, therefore,

x+y≥20

15x+10y≤270 and x+y≥20 are the constraints for this situation.

Keywords: inequality, addition

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Answer:

a)  Right-tailed test

b) 0.006569

c) We reject [tex]H_0[/tex].

Step-by-step explanation:

We are given the following information in the question:

[tex]z_{\text{statistic}} = 2.48[/tex]

We are testing the claim  that p > 0.7

Alpha, α = 0.05

a) We have to identify type of test.

Since we test for claim p > 0.7, it is a one-tailed(right) test.

b) The p-value can be calculated with the help of standard normal table.

[tex]P(z>2.48) = 1 - P(z<2.48) = 0.006569[/tex]

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

Thus, we reject [tex]H_0[/tex].

The hypothesis test in question is right-tailed. The p-value can be obtained from a Z-table or statistical software and is probably less than 0.01. Given a significance level of alpha=.05, we should reject H0 if the p-value is less than alpha.

This question involves the concepts of hypothesis testing, p-value, and significance level which are typically covered in a college-level statistics course.

(a) The test is right-tailed because we are testing the claim that p > 0.7. A right-tailed test looks at whether the test statistic is greater than the hypothesized population parameter.

(b) To find the p-value, you would need to look up the value of z=2.48 in a standard normal (Z) table or use statistical software. The p-value represents the probability that the observed difference occurred by chance assuming the null hypothesis is true. The exact value would depend on your table or software, but generally, a z-score of 2.48 would yield a very small p-value, probably less than 0.01.

(c) If the p-value is less than your significance level (alpha=.05), you would reject the null hypothesis (H0). That happens because the evidence suggests that it is highly unlikely (less than a 5% chance) that the observed difference occurred by chance.

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Answer:

The correct option is e) [tex]H_0: \mu_1=\mu_2\ and\ H_a:\mu_1\neq \mu_2[/tex]

Step-by-step explanation:

Consider the provided information.

Ten college students reported that they consume coffee but do not smoke cigarettes; these comprise the Coffee-Only group, or Group 1. Another ten college students reported that they consumed coffee and also smoked cigarettes; these comprise the Coffee + Cigarettes group, or Group 2.

The null hypothesis tells the population parameter is equal to the claimed value.

If there is no statistical significance in the test then it is know as the null which is denoted by [tex]H_0[/tex], otherwise it is known as alternative hypothesis which denoted by [tex]H_a[/tex].

The amount of coffee that  "coffee drinkers" consume on a weekly basis differs depending on whether the coffee drinker also smokes cigarettes.

Thus, the required hypothesis are: [tex]H_0: \mu_1=\mu_2\ and\ H_a:\mu_1\neq \mu_2[/tex]

Therefore, the correct option is e) [tex]H_0: \mu_1=\mu_2\ and\ H_a:\mu_1\neq \mu_2[/tex]

Answer:

Null hypothesis:[tex]p\leq 0.79[/tex]  

Alternative hypothesis:[tex]p > 0.79[/tex]  

[tex]z=\frac{0.849 -0.79}{\sqrt{\frac{0.79(1-0.79)}{370}}}=2.786[/tex]  

[tex]p_v =P(z>2.786)=0.00267[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/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 5% of significance the true proportion is higher than 0.79.  

Step-by-step explanation:

1) Data given and notation

n=750 represent the random sample taken

X=314 represent the respondents that use social media in their job search.

[tex]\hat p=\frac{314}{370}=0.849[/tex] estimated proportion of respondents that use social media in their job search

[tex]p_o=0.79[/tex] is the value that we want to test

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

Confidence=95% or 0.95

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 the true proportion is hgiher than 0.79.:  

Null hypothesis:[tex]p\leq 0.79[/tex]  

Alternative hypothesis:[tex]p > 0.79[/tex]  

When we conduct a proportion test we need to use the z statisticc, 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{0.849 -0.79}{\sqrt{\frac{0.79(1-0.79)}{370}}}=2.786[/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=0.05[/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>2.786)=0.00267[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/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 5% of significance the true proportion is higher than 0.79.  

Answer:

112

Step-by-step explanation:

Let A be a subset of S that satisfies such condition.

If 3∈A, then the other three elements of A must be chosen from the set B={1,2,5,6,7,8,9,10} (because 3 cannot be chosen again and 4 can't be alongside 3). B has eight elements, then there are [tex]\binom{8}{3}=56[/tex] ways to select the remaining elements of A (the binomial coefficient counts this). The remaining elements determine A uniquely, then there are 56 subsets A.

If 4∉A, we have to choose the remaining elements of A from the set B={1,2,5,6,7,8,9,10}. B has eight elements, then there are [tex]\binom{8}{3}=56[/tex] ways to select the remaining elements of A. Thus, there are 56 choices for A.

By the sum rule, the total number of subsets is 56+56=112

Answer:

a. gender (two levels: male, female)

d. class size (three levels: small, medium, large)

Step-by-step explanation:

The study is going to be made to identify how male and female students react to the class size(small, medium, large). So there are two correct options, since the gender and the class size are factors in this research.

The correct options are:

a. gender (two levels: male, female)

d. class size (three levels: small, medium, large)

In this example, the factors and levels are gender, academic performance, and class size.

The factors and levels in this example are:

a. Gender (two levels: male, female)

b. Academic performance (three levels: above average, average, below average)

c. Class size (two levels: small, large)

Therefore, the answer is a, b, and c.

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The petrol consumption for David's journey, considering bounds, is between 6.5439 km/l and 6.8182 km/l.

To determine the petrol consumption (c) for David's journey, we use the formula c = d/p, where d is the distance (187 km) and p is the amount of petrol used (28 litres).

Firstly, we calculate the maximum and minimum possible values for c by considering the upper and lower bounds of d and p:

Maximum value of c:

Upper bound of distance (d) = 187 + 0.5 (half of the smallest decimal place in 187 km) = 187.5 km

Lower bound of petrol used (p) = 28 - 0.5 (half of the smallest decimal place in 28 litres) = 27.5 litres

Maximum c = (187.5 km) / (27.5 litres) = 6.8182 km/l

Minimum value of c:

Lower bound of distance (d) = 187 - 0.5 = 186.5 km

Upper bound of petrol used (p) = 28 + 0.5 = 28.5 litres

Minimum c = (186.5 km) / (28.5 litres) = 6.5439 km/l

Therefore, considering bounds, the petrol consumption (c) for David's journey is between 6.5439 km/l and 6.8182 km/l.

Answer:

We can conclude that  the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997 (P-value=0.00014).

Step-by-step explanation:

We have to perform an hypothesis test on a proportion.

The null and alternative hypothesis are:

[tex]H_0: \pi=0.46\\\\H_1: \pi>0.46[/tex]

The significance level is α=0.05.

The standard deviation is estimated as:

[tex]\sigma=\sqrt{\frac{\pi(1-\pi)}{N} } =\sqrt{\frac{0.46(1-0.46)}{1010} }=0.0157[/tex]

The z value for this sample is

[tex]z=\frac{p-\pi-0.5/N}{\sigma} =\frac{525/1010-0.46-0.5/1010}{0.0157} =\frac{0.52-0.46-0.00}{0.0157}=\frac{0.06}{0.0157} =3.822[/tex]

The P-value for z=3.822 is P=0.00014.

The P-value is smaller than the significance level, so the effect is significant. The null hypothesis is rejected.

We can conclude that  the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997.

Answer:

X=2

Step-by-step explain

4\div 10(12-3x)=3\div 10(12x-16)

when we do cross multiply then we get

40(12-3x)=30(12x-16)

we can divide by 10 both side and solve then we get

48-12x=36x-48

and after solving this equation we get

x=2

Answer:

a) 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b) 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c) 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].

Normal probability distribution

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.

In this problem, we have that:

[tex]\mu = 9.6, \sigma = 0.5[/tex].

a. Find the probability that a randomly selected woman has a foot length less than 10.0 in

This probability is the pvalue of Z when [tex]X = 10[/tex].

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

[tex]Z = \frac{10 - 9.6}{0.5}[/tex]

[tex]Z = 0.8[/tex]

[tex]Z = 0.8[/tex] has a pvalue of 0.7881.

So there is a 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b. Find the probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 8.

When X = 10, Z has a pvalue of 0.7881.

For X = 8:

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

[tex]Z = \frac{8 - 9.6}{0.5}[/tex]

[tex]Z = -3.2[/tex]

[tex]Z = -3.2[/tex] has a pvalue of 0.0007.

So there is a 0.7881 - 0.0007 = 0.7874 = 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c. Find the probability that 25 women have foot lengths with a mean greater than 9.8 in.

Now we have [tex]n = 25, s = \frac{0.5}{\sqrt{25}} = 0.1[/tex].

This probability is 1 subtracted by the pvalue of Z when [tex]X = 9.8[/tex]. So:

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

[tex]Z = \frac{9.8 - 9.6}{0.1}[/tex]

[tex]Z = 2[/tex]

[tex]Z = 2[/tex] has a pvalue of 0.9772.

There is a 1-0.9772 = 0.0228 = 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

Answer:

3. What is the probability that an adult selected at random has both a landline and a cell phone?

A. 0.58

4. Given an adult has a cell phone, what is the probability he does not have a landline?

C. 0.3012

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that an adult has a landline at his residence.

B is the probability that an adult has a cell phone.

C is the probability that a mean is neither of those.

We have that:

[tex]A = a + (A \cap B)[/tex]

In which a is the probability that an adult has a landline but not a cell phone and [tex]A \cap B[/tex] is the probability that an adult has both of these things.

By the same logic, we have that:

[tex]B = b + (A \cap B)[/tex]

The sum of all the subsets is 1:

[tex]a + b + (A \cap B) + C = 1[/tex]

2% of adults have neither a cell phone nor a landline.

This means that [tex]C = 0.02[/tex].

73% of adults have a landline at their residence (event A); 83% have a cell phone (event B)

So [tex]A = 0.73, B = 0.83[/tex].

What is the probability that an adult selected at random has both a landline and a cell phone?

This is [tex]A \cap B[/tex].

We have that [tex]A = 0.73[/tex]. So

[tex]A = a + (A \cap B)[/tex]

[tex]a = 0.73 - (A \cap B)[/tex]

By the same logic, we have that:

[tex]b = 0.83 - (A \cap B)[/tex].

So

[tex]a + b + (A \cap B) + C = 1[/tex]

[tex]0.73 - (A \cap B) + 0.83 - (A \cap B) + (A \cap B) + 0.02 = 1[/tex]

[tex](A \cap B) = 0.75 + 0.83 - 1 = 0.58[/tex]

So the answer for question 3 is A.

4. Given an adult has a cell phone, what is the probability he does not have a landline?

83% of the adults have a cellphone.

We have that

[tex]b = B - (A \cap B) = 0.83 - 0.58 = 0.25[/tex]

25% of those do not have a landline.

So [tex]P = \frac{0.25}{0.83} = 0.3012[/tex]

The answer for question 4 is C.

Answer:

The recursive formula for given series is: [tex]a_{n} = a_{n-1} + 4[/tex]

The explicit formula is given by: [tex]a_{n} = 10 + 5(n-1)[/tex]

Step-by-step explanation:

Determining the Recursive Formula:

         d = 6 -2 = 10 - 6 = 14 - 10 = 4

Determining the Explicit  Rule or Formula:

An explicit formula defines the nth term of the sequence, where n being the term's location. In other words, a sequence can be defined as a formula in  terms of n. So,

Hence, the explicit formula is given by:

where,

aₙ is the nth term of the sequence

n is the term number

a₁ is the first term

d is the common difference

As the given sequence is:  2,6,10,14

a₁ = first term = 2

d = common difference = 6 - 2 = 4

Using explicit formula:

[tex]a_{n} = a_{1} + d (n-1)[/tex]

[tex]a_{n} = 10 + 5(n-1)[/tex]

[tex]a_{n} = 10 + 5n-5[/tex]

[tex]a_{n} = 5n+ 5[/tex]

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Final answer :

The recursive rule for the sequence is Sn =  Sn- 1 4 with S1 =  2, and the  unequivocal rule is Sn =  4n- 2. Both rules help in determining the sequence's terms either by using the  former term or directly calculating any term's value.  

Explanation:

To determine the recursive rule and the  unequivocal rule for the sequence 2, 6, 10, 14, we need to observe the pattern of the sequence and  produce formulas that can  induce the terms of the sequence. This sequence is  computation because the difference between each term is constant.

Recursive Rule

A recursive rule for a sequence defines each term grounded on the  former term. For the given sequence, each term after the first is  attained by adding 4 to the  former term. The recursive rule can be written as  Sn =  Sn- 1 4, for n> 1  where S1 =  2 is the first term.  

unequivocal Rule

An  unequivocal rule, also known as a unrestricted- form expression, defines the  utmost term of a sequence as a function of n without  demanding the  former term. For this  computation sequence, the  unequivocal rule can be determined by using the formula Sn =  a( n- 1) d, where' a' is the first term and'd' is the common difference. The  unequivocal rule for our sequence is  Sn =  2( n- 1) × 4 =  4n- 2.

Answer:

Because the​ P-value is _(0.02)  less than the significance level 0.05​, there is sufficient evidence to support the claim that there is a linear correlation between lemon imports and crash fatality rates for a significance level of α=0.05.

C. The results suggest that imported lemons cause car fatalities.

Step-by-step explanation:

Hello!

The study variables are:

X₁: Weight of imported lemons.

X₂: Car crash fatality rate.

The objective is to test if the imported lemons affect the occurrence of car fatalities. To do so you are asked to use a linear correlation test.

I've made a Scatterplot with the given data, it is attached to the answer.

To be able to use the parametric linear correlation you can use the parametric test (Person) or the nonparametric test Spearman. For Person, you need your variables to have a bivariate normal distribution. Since one of the variables is a discrete variable (ratio of car crashes) and the sample is way too small to make an approximation to a normal distribution, the best test to use is Spearman's rank correlation.

This correlation coefficient (rs) takes values from -1 to 1

If rs = -1 this means that there is a negative correlation between the variables

If rs= 1 this means there is a positive correlation between the variables

If rs =0 then there is no correlation between the variables.

The hypothesis is:

H₀: There is no linear association between X₁ and X₂

H₁: There is a linear association between X₁ and X₂

α: 0.05

To calculate the Spearman's correlation coefficient you have to assign ranks to the observed values of each variable, from the smallest to the highest). Then you have to calculate the difference (d)between the ranks and the square of that difference (d²). (see attachment)

The formula for the correlation coefficient is:

[tex]rs= 1 - \frac{6* (sum of d^2)}{(n-1)n(n+1)}[/tex]

[tex]rs= 1 - \frac{6* (40)}{4*5*6}[/tex]

rs= -1

For this value of the correlation coefficient, the p-value is 0.02

Since the p-value (0.02) is less than the significance level (0.05) the decision is to reject the null hypothesis. In other words, there is a linear correlation between the imported lemons and the car crash fatality ration, this means that the modification in the lemon import will affect the car crash fatality ratio.

Note: the correlation coefficient is negative, so you could say that there is a correlation between the variables and this is negative (meaning that when the lemon import increases, the car crash fatality ratio decreases)

I hope it helps!

Answer:

Probability = 0.100.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Eithey they have one or more of these conditions. Or they do not. This means that the binomial probability distribution is used to solve this problem.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

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.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]n = 750, p = 0.45[/tex]

So

[tex]E(X) = 750*0.45 = 337.5[/tex]

[tex]\sqrt{Var(X)} = \sqrt{750*0.45*0.55} = 13.62[/tex]

Calculate the probability that fewer than 320 out of the n = 750 adults over 65 in the study suffer from one or more of the conditions under consideration.

This is the pvalue of Z when [tex]X = 320[/tex]. So:

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

[tex]Z = \frac{320 - 337.5}{13.62}[/tex]

[tex]Z = -1.28[/tex]

[tex]Z = -1.28[/tex] has a pvalue of 0.100.

So

Probability = 0.100.

Answer:

a) [tex]y=-6.254 x +75.064[/tex]  

b) r =-0.932

The % of variation is given by the determination coefficient given by [tex]r^2[/tex] and on this case [tex]-0.932^2 =0.8687[/tex], so then the % of variation explained by the linear model is 86.87%.

Step-by-step explanation:

Assuming the following dataset:

Monthly Sales (Y)     Interest Rate (X)

       22                               9.2

       20                               7.6

       10                                10.4

       45                                5.3

Part a

And we want a linear model on this way y=mx+b, where m represent the slope and b the intercept. In order to find the slope we have this formula:

[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]  

Where:  

[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]  

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]  

With these we can find the sums:  

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=278.65-\frac{32.5^2}{4}=14.5875[/tex]  

[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=696.9-\frac{32.5*97}{4}=-91.225[/tex]  

And the slope would be:  

[tex]m=\frac{-91.225}{14.5875}=-6.254[/tex]  

Nowe we can find the means for x and y like this:  

[tex]\bar x= \frac{\sum x_i}{n}=\frac{32.5}{4}=8.125[/tex]  

[tex]\bar y= \frac{\sum y_i}{n}=\frac{97}{4}=24.25[/tex]  

And we can find the intercept using this:  

[tex]b=\bar y -m \bar x=24.25-(-6.254*8.125)=75.064[/tex]  

So the line would be given by:  

[tex]y=-6.254 x +75.064[/tex]  

Part b

For this case we need to calculate the correlation coefficient given by:

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]  

[tex]r=\frac{4(696.9)-(32.5)(97)}{\sqrt{[4(278.65) -(32.5)^2][4(3009) -(97)^2]}}=-0.937[/tex]  

So then the correlation coefficient would be r =-0.932

The % of variation is given by the determination coefficient given by [tex]r^2[/tex] and on this case [tex]-0.932^2 =0.8687[/tex], so then the % of variation explained by the linear model is 86.87%.

Answer:

a) Q₀ × 1.335 cps

b) 12 fourths

c)  5 octaves

Step-by-step explanation:

The circle of fourth is generated by starting at any note and stepping upward by intervals of a fourth(five half-steps)

(a)  By what factor is the frequency of a tone increased if it is raised by a​ fourth?

consider the following exponential growth formula :

Q = Q₀ × f °

Q = Q₀ × 1.05946 ⁿ

Substituting n = 5

Q = Q₀ × 1.05946 ⁿ

Q = Q₀  × 1.05946⁵

Q = Q₀  × 1.335 cps

Therefore, note that  five half-step higher will be increased by factor 1.335

(b)  How many fourths are required to complete the entire circle of​ fourths?

SOLUTION

In each step, number of half steps = 5

Total number of half steps in one octave = 12 half- steps

Therefore, total number of fourths required to complete the entire circle of fourths = 12 fourths

(c) Total number of half steps in one octave = 12 half-steps  

Total half steps in complete circle of fourths  

= 12×5  

= 60 half-steps

Calculating number of octaves (dividing it by 12)  

= 60/12  

= 5 octaves

It is given that the circle of fourth is generated by starting at any note and stepping upward by intervals of a fourth(five half-steps)

[tex]Q = Q_0 \times f ^0\\Q = Q_0 \times 1.05946^n[/tex]

Put n = 5 in the above formula, we get

[tex]Q = Q_0 ( 1.05946 ^n)\\Q = Q_0 ( 1.05946 ^5)\\Q = Q_0 (1.335) cps[/tex]

So, by 1.335  factor the frequency of a tone increased if it is raised by a​ fourth.

Total number of half steps in one octave = 12 half- steps

So, total number of fourths required to complete the entire circle of fourths = 12 fourths

Total half steps in complete circle of fourths will be  12×5  = 60 half-steps

Now, the number of octaves is [tex]\frac{60}{12} =5[/tex]  octaves.

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Answer:

a) H0=Weight of detergent=500gr  (null hypotheses )

H1=Weight of detergent>500 gr. (alternative hypotheses )

b) X test.

Step-by-step explanation:

a)

The statement from which we are going to establish the assumptions is that supplied by the manufacturer, which says that the average weight of the detergent is 500 g. So:

H0=Weight of detergent=500gr  (null hypotheses )

H1=Weight of detergent>500 gr. (alternative hypotheses )

b)

You are not providing the source you mention where the sample of detergent boxes taken is located. So I will explain in general terms how the problem should be posed.

For this case, we can use Z test because the sample we are taking is larger than 30.

The first thing we have to do is calculate Z, which is given by:

[tex]Z=\frac{(X-Xav)}{SD}[/tex]

Where:

Xav=Average of x.

SD=Standard deviation.

Then, we look for the table and if the value we have just calculated is greater than that of the table, then we must reject the null hypothesis.

Answer:

D. All of the above

Step-by-step explanation:

By the given data, some persons were in the experiment and each of them was given with three meals, Low Calorie, Moderate Calorie and High Calorie. They results for the meal they liked is as below:

Low Calorie Meal got likes = 6

Moderate Calorie Meal got likes = 7

High Calorie Meal got likes = 17

so,

Option A is correct as High Calorie meal got 17 likes while Low Calorie meal got 6 likes.

Option B is also correct as Low Calorie meal got 6 likes while Moderate Calorie meal got 7 likes.

Option C is correct too as High Calorie meal got largest number of likes even more than the double of Low calorie and Moderate calorie meal so it was more than expected.

Final answer:

Based on the data provided, without performing an actual chi-squared test due to a lack of p-value or statistical data, the most supported conclusion is that participants liked the high calorie meal more than was expected (option C). This is inferred from the observed preference for the high calorie meal, which was chosen significantly more often (17 times) compared to its expected frequency (10 times).

Explanation:

The question presents a scenario where participants are asked to taste three types of meals with different calorie counts and choose the one they like best. The observed frequencies (fo) for each type of meal (low calorie, moderate calorie, and high calorie) are given as 6, 7, and 17, respectively. The expected frequencies (fe) are all 10, assuming no preference among the meals. Using a significance level of .05, we must employ a chi-squared test to determine if the observed preferences significantly differ from what was expected given no effect of calorie count.

The chi-squared test result for this would involve calculating the sum of squared differences between observed and expected frequencies, divided by the expected frequencies for each category and summing those values. However, since the data on the actual chi-squared statistic or the p-value are not provided, we cannot perform the exact calculation. Nonetheless, we can make a conclusion based on the observed and expected frequencies:

There is a notable preference for the high calorie meal, as it was chosen substantially more often than expected (17 vs. 10).

The low calorie meal was chosen less often than expected (6 vs. 10).

The moderate calorie meal was also chosen less than expected, but to a lesser extent (7 vs. 10).

Without a specific p-value or chi-squared statistic, we can't formally conclude that the participants liked the high calorie meal significantly more than the other meals, but the observed data suggest that might be the case. Also, we cannot definitively reject the possibility that the preference for the high-calorie meal is due to chance.

Therefore, the most reasonable conclusion based on the given information without formal test statistics would be option C: Participants liked the high calorie meal more than was expected.

This is in line with the general principle that a higher calorie meal tends to be preferred potentially due to higher fat content, which has been associated with better taste and satisfaction, particularly in fast-food meals. At a .05 level of significance and without the actual test statistic, option C is the most supported by the provided data.

Answer:

option (d) 800,000

Step-by-step explanation:

Let the number of pairs that must be sold be 'x'

Thus,

Total variable cost for x units = $5x

Therefore,

Total cost = Fixed cost + Total variable cost

= $6 million + $5x

Now,

Profit = Cost - Revenue

Thus,

$2 million = $15x - ( $6 million + $5x )

or

$2 million + $6 million = $10x

or

$8 million = $10x

or

$8,000,000 = $10x

or

x = 800,000

Hence,

option (d) 800,000

M^2 + MNP

(3)^2 + (3)(4)(7)

9 + 12(7)

9 + 84

93

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Answer:

1.83%

Step-by-step explanation:

For the expected value of 4 cases per 1000 children. We can use a formula for Poisson distribution to calculate the probability that none of the 1000 students would contract the flu:

[tex]P(X = 0) = \frac{\lambda^ke^{-\lambda}}{k!}[/tex]

[tex]P(X = 0) = \frac{4^0 e^{-4}}{0!}[/tex]

[tex]P(X = 0) = \frac{1}{e^4} = 0.0183 [/tex]

So the probability of this to happens is 1.83%

The probability that none of the 1,000 children will contract the flu after getting vaccinated, given the average number of flu cases after vaccination is approximately 0.0183 or 1.83%, using the Poisson distribution.

In this question, we are required to use the Poisson distribution to find the probability that none of the 1,000 children will contract the flu this season, given that the average number of cases per 1,000 children after vaccination is 4 (λ = 4).

A Poisson probability can be calculated using the formula P(x; λ) = (e^-λ * λ^x) / x!, where x represents the actual number of successes that result from the experiment, e is the base of the natural logarithm approximated to 2.71828, and λ is the mean number of successes that occur in a specified region.

To find the probability that zero children contract the flu, we use x = 0 in the formula. The Poisson probability P(0; 4) = ((e^-4 * 4^0) / 0!) = e^-4 = 0.0183.

So, the probability that none of the 1,000 children will contract the flu after getting vaccinated is approximately 0.0183 or 1.83%.

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Answer:

1) [tex]m=\frac{77}{10.5}=7.333[/tex]

2) [tex]b=\bar y -m \bar x=80.857-(7.333*3.5)=55.190[/tex]

3) [tex]\hat y=7.333(0)+55.190=55.190[/tex]

4) False. The values predited will fall on the same line since we are estimating  the values with just one line.

5) [tex]\hat y=7.333(4.5)+55.190=88.190[/tex]

6) [tex]R^2 = (0.971^2) =0.943[/tex]

And that means that the linear model explains 94.29% of the variation.

Step-by-step explanation:

We assume that the data is this one:

x:1.5,2.5, 3, 3.5 , 4, 4.5 ,5.5

y: 67, 69, 80, 81, 86, 89, 94.

Step 1 of 6: Find the estimated slope. Round your answer to three decimal places.

For this case we need to calculate the slope with the following formula:

[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]

Where:

[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]

So we can find the sums like this:

[tex]\sum_{i=1}^n x_i = 1.5+2.5+3+3.5+4+4.5+5.5=24.5[/tex]

[tex]\sum_{i=1}^n y_i =67+ 69+ 80+ 81+ 86+ 89+ 94=566[/tex]

[tex]\sum_{i=1}^n x^2_i =1.5^2+2.5^2+3^2+3.5^2+4^2+4.5^2+5.5^2=96.25[/tex]

[tex]\sum_{i=1}^n y^2_i =67^2+69^2+80^2+81^2+86^2+89^2+94^2=46364[/tex]

[tex]\sum_{i=1}^n x_i y_i =1.5*67+2.5*69+3*80+3.5*81+4*86+4.5*89+5.5*94=2058[/tex]

With these we can find the sums:

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=96.25-\frac{24.5^2}{7}=10.5[/tex]

[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}=2058-\frac{24.5*566}{7}=77[/tex]

And the slope would be:

[tex]m=\frac{77}{10.5}=7.333[/tex]

Step 2 of 6: Find the estimated y-intercept. Round your answer to three decimal places.

Now we can find the means for x and y like this:

[tex]\bar x= \frac{\sum x_i}{n}=\frac{24.5}{7}=3.5[/tex]

[tex]\bar y= \frac{\sum y_i}{n}=\frac{566}{7}=80.857[/tex]

And we can find the intercept using this:

[tex]b=\bar y -m \bar x=80.857-(7.333*3.5)=55.190[/tex]

So the line would be given by:

[tex]\hat y=7.333 x +55.190[/tex]

Step 3 of 6: Determine the value of the dependent variable yˆ at x = 0.

[tex]\hat y=7.333(0)+55.190=55.190[/tex]

Step 4 of 6: Determine if the statement "Not all points predicted by the linear model fall on the same line" is true or false.

False. The values predited will fall on the same line since we are estimating  the values with just one line.

Step 5 of 6: Find the estimated value of y when x = 4.5. Round your answer to three decimal places.

[tex]\hat y=7.333(4.5)+55.190=88.190[/tex]

Step 6 of 6: Find the value of the coefficient of determination. Round your answer to three decimal places

n=7 [tex] \sum x = 24.5, \sum y = 566, \sum xy =2058, \sum x^2 =96.25, \sum y^2 =46364[/tex]  

And in order to calculate the correlation coefficient we can use this formula:

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]

[tex]r=\frac{7(2058)-(24.5)(566)}{\sqrt{[7(96.25) -(24.5)^2][7(46364) -(566)^2]}}=0.971[/tex]

And the determination coeffcient is just the square of the correlation coefficient given by:

[tex]R^2 = (0.971^2) =0.943[/tex]

And that means that the linear model explains 94.3% of the variation.

Regression analysis can be done to find the relationship between hours spent studying and midterm exam grades using the regression line equation yˆ = b0 + b1x. But doing this requires statistical computations and ensuring the correlation coefficient is statistically significant. The coefficient of determination, which indicates the proportion of the variance in the dependent variable predictable from the independent variable, is also needed.

The equation of the regression line you're referring to is written as yˆ = b0 + b1x, where yˆ is the predicted value of the dependent variable (in this case, Midterm Grades), b0 is the y-intercept, b1 is the slope or regression coefficient, and x is the value of the independent variable (Hours Studying). We can use the given data to find the estimated slope and y-intercept, value of yˆ at x = 0, test the truthfulness of the statement, and find the estimated value of y when x = 4.5.

However, being able to do this requires the use of statistical software or methods to calculate corresponding values. Also, this also requires us to know the value of the correlation coefficient to determine if it's statistically significant, as it's not wise to make predictions using a regression line with a correlation coefficient that's not statistically significant.

The last piece to find is the value of the coefficient of determination, which shows the proportion of the dependent variable's variance that's predictable from the independent variable(s). This value also needs computation using formulas or statistical software.

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Answer:

There is no evidence that there is no significant difference between the sample means

Step-by-step explanation:

given that a  statistics instructor who teaches a lecture section of 160 students wants to determine whether students have more difficulty with one-tailed hypothesis tests or with two-tailed hypothesis tests. On the next exam, 80 of the students, chosen at random, get a version of the exam with a 10-point question that requires a one-tailed test. The other 80 students get a question that is identical except that it requires a two-tailed test. The one-tailed students average 7.81 points, and their standard deviation is 1.06 points

The two-tailed students average 7.64 points, and their standard deviation is 1.33 points.

Group   One tailed X     Two tailed Y  

Mean 7.8100 7.6400

SD 1.0600 1.3300

SEM 0.1185 0.1487

N 80       80  

[tex]H_0:\bar x=\bar y\\H_a: \bar x \neq \bar y[/tex]

(Two tailed test)

The mean of One tailed X minus Two tailed Y equals 0.1700

t = 0.8940

 df = 158

p value =0.3727

 p is greater than alpha 0.05

There is no evidence that there is no significant difference between the sample means

Answer:

We conclude that the  bag filling machine works correctly at the 444.0 gram setting.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 444.0 gram

Sample mean, [tex]\bar{x}[/tex] = 443.0 grams

Sample size, n = 40

Alpha, α = 0.02

Population standard deviation, σ = 23.0 grams

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 444.0\text{ grams}\\H_A: \mu < 444.0\text{ grams}[/tex]

We use one-tailed(left) 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{443 - 444}{\frac{23}{\sqrt{40}} } =-0.274[/tex]

Now, [tex]z_{critical} \text{ at 0.02 level of significance } = -2.054[/tex]

Since,  

[tex]z_{stat} < z_{critical}[/tex]

We fail to reject the null hypothesis and accept the null hypothesis. Thus, we conclude that the  bag filling machine works correctly at the 444.0 gram setting.

The decision rule for the hypothesis test to check machine calibration involves finding the z-value for a sample mean of 443.0 grams and comparing it with the critical z-value corresponding to a 0.02 significance level for a left-tailed test. If the z-value is less than the critical value, the null hypothesis is rejected.

To determine the decision rule for testing whether the bag filling machine is correctly set at 444.0 grams when we suspect underfilling, we need to perform a hypothesis test using the z-test since the standard deviation is known. Given the sample size (n = 40), sample mean (μ = 443.0 grams), population mean (μ0 = 444.0 grams), population standard deviation (σ = 23.0 grams), and a level of significance of 0.02, we can formulate the null hypothesis (H0: μ = μ0) and the alternative hypothesis (H1: μ < μ0).

The decision rule involves comparing the computed z-value to the critical value from the standard normal distribution at the 0.02 level of significance for a left-tailed test. If the computed z-value is less than the critical z-value, we reject the null hypothesis and accept that the machine is underfilling. The critical z-value is found using the z-table, which corresponds to a cumulative probability of 0.02.

Answer:

Null hypothesis:[tex]\mu \geq 4.0[/tex]    

Alternative hypothesis:[tex]\mu < 4.00[/tex]    

[tex]t=\frac{3.48-4.00}{\frac{1.150075}{\sqrt{45}}}=-3.033077[/tex]    

[tex]df=n-1=45-1=44[/tex]

Since is a left-sided test the p value would be:    

[tex]p_v =P(t_{(44)}<-3.033077)=0.002025[/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 reject the null hypothesis, so we can conclude that the mean is significantly less than 4.00 at 5% of significance.  

Step-by-step explanation:

Data given and notation    

[tex]\bar X=3.48[/tex] represent the sample mean

[tex]s=1.150075[/tex] represent the sample standard deviation    

[tex]n=45[/tex] sample size    

[tex]\mu_o =4.00[/tex] represent the value that we want to test  

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.    

z 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.    

We need to conduct a hypothesis in order to check if the mean is less than 4.00 :    

Null hypothesis:[tex]\mu \geq 4.0[/tex]    

Alternative hypothesis:[tex]\mu < 4.00[/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{3.48-4.00}{\frac{1.150075}{\sqrt{45}}}=-3.033077[/tex]    

P-value    

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

[tex]df=n-1=45-1=44[/tex]

Since is a left-sided test the p value would be:    

[tex]p_v =P(t_{(44)}<-3.033077)=0.002025[/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 reject the null hypothesis, so we can conclude that the mean is significantly less than 4.00 at 5% of significance.    

Answer:

Step-by-step explanation:

Answer:

The equation would be [tex]30=12+9x[/tex].

Allen will Run for 2 hours.

Step-by-step explanation:

Given:

Total Distance traveled = 30 miles

Average rate of walking = 3 mi/hr

Average rate of running = 9 mi/hr

Time for walking = 4 hours

W need to find the time for running.

Solution:

Let Number of hrs required for running be 'x'.

Total Distance traveled is equal Distance Traveled in walking plus Distance traveled in Running

But Distance is equal Rate multiplied by Time.

Framing in equation form we get;

Hence Total Distance Traveled = Rate of walking × Hours of walking + Rate of Running × Hours of running.

Substituting the values we get;

[tex]30 = 3\times4 +9x\\\\30=12+9x[/tex]

Hence The equation would be [tex]30=12+9x[/tex]

Solving above equation we get;

[tex]9x=30-12\\\\9x=18\\\\x=\frac{18}{9}=2\ hrs[/tex]

Hence Allen will Run for 2 hours.

Answer:

A) [tex]P(t)=\frac{2000}{e^{ln4e^{-0.1t}}}[/tex]

B) P(t→∞)=2000

C) [tex]P=\frac{K}{e}=\frac{carrying capacity}{e}[/tex]

Step-by-step explanation:

Given differential eq is

                      [tex]\frac{dP}{dt}=c ln (\frac{K}{P})P[/tex] --- (1)

Eq is separable

                     [tex]\frac{1}{ln (\frac{K}{P})P}dP=cdt[/tex] --- (2)

                     [tex]let \\u = ln\frac{K}{P}\\du= \frac{1}{\frac{K}{P}}(\frac{-K}{P^{2}}).dP\\du=\frac{-1}{P}.dP\\dP=-P.du[/tex]

substituting in  (2)

[tex]-\frac{du}{u}=dt[/tex]

Integrating both sides

[tex]\int {-\frac{1}{u}} \, du=\int{c}\,dt\\-ln|u|=ct +B\\ln|u|=-ct -B\\[/tex]

Back substituting value of u

[tex]ln |ln\frac{K}{P}|=-ct-B\\|ln\frac{K}{P}|=e^{-ct-B}\\ln|\frac{K}{P}|=be^{-ct}\\[/tex]---(3)

at t =0

[tex]ln|\frac{K}{P}|=be^{-ct}\\b=ln|\frac{K}{P}|\\b=ln\frac{2000}{500}\\b=ln|4|[/tex]

from (3)

[tex]ln|\frac{K}{P}|=be^{-ct}\\\frac{K}{P}=e^{ln4e^{-ct}}\\P(t)=\frac{K}{e^{ln4e^{-ct}}}[/tex]

[tex]P(t)=\frac{2000}{e^{ln4e^{-0.1t}}}[/tex]

B) [tex]\lim{t \to \infty}[/tex]

[tex]P( {t \to \infty} )=\frac{2000}{e^{ln4e^{-0.1\infty}}}\\e^{-0.1\infty}=0\\\implies P( {t \to \infty} )=\frac{2000}{e^{0}}}\\\\P(\infty)=2000\\[/tex]

which is the carrying capacity.

C) To find the fastest growth rate we have to maximize [tex]\frac{dP}{dt}[/tex]

From given differential eq

[tex]\frac{dP}{dt}=cln|\frac{K}{P}|P[/tex]

so function to maximize is

[tex]f(P)=cln|\frac{K}{P}|P[/tex]

[tex]f'(P)=cln|\frac{K}{P}|+c\frac{1}{\frac{K}{P}}\frac{-K}{P^{2}}.P[/tex]

[tex]f'(P)=c[ln|\frac{K}{P}|-1][/tex]

To maximize find f'(P)=0

[tex]c[ln|\frac{K}{P}|-1]=0[/tex]

[tex]ln|\frac{K}{P}|=1[/tex]

[tex]\frac{K}{P}=e[/tex]

[tex]P=\frac{K}{e}=\frac{carrying capacity}{e}[/tex]

The Gompertz function models population growth considering the carrying capacity. solve the Gompertz differential equation for c=0.1, K=2000, P0=500. The carrying capacity (K) gives the limiting value of the population size (2000). The time at which the population growth is fastest can be calculated by taking the derivative of the population function, setting it to zero and solving for P.

The Gompertz function is a model of growth that was developed to model population growth, considering the factor of carrying capacity. Instead of compound exponential growth, it models the growth as slowing down as it reaches the limit of the carrying capacity. In your particular case, solving the differential equation for the variables provided (c=0.1, K=2000, P0=500) would require the integration techniques and use of logarithmic functions.

(a): The full solution for the Gompertz function involves advanced mathematics, you should expect some intricate function in the form of P(t) = ... The particulars depend on the specifics of the integration process.

(b): The limiting value of the population size as t→infinity (limt→[infinity]P(t) will be K, which in this instance equals 2000. This is due to the concept of carrying capacity. Beyond this value, the environment/conditions can no longer support additional growth.

(c): Finding the time at which population growth is fastest involves setting the derivative of the population function to zero and solving for P. The solution P=... can be calculated using standard techniques of calculus.

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The price of one ticket is $ 62.2

Given that a performer expects to sell 5000 tickets for an upcoming event

They want to make a total of $ 311, 000 in sales from these tickets

To find: price of one ticket

Let us assume that all tickets have the same price

Let "a" be the price of one ticket

So the total sales price of $ 311, 000 is obtained from product of 5000 tickets and price of one ticket

[tex]\text {total sales price }=5000 \times \text { price of one ticket }[/tex]

[tex]311000 = 5000 \times a\\\\a = \frac{311000}{5000}\\\\a = 62.2[/tex]

Thus the price of one ticket is $ 62.2

Answer:

Critical value: [tex]z= 2.575[/tex]

99% confidence interval: (0.695 cc/cubic meter, 0.713 cc/cubic meter).

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex] is the critical value

Now, find M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}} = 2.575*\frac{0.0142}{\sqrt{17}} = 0.0089[/tex]

The lower end of the interval is the mean subtracted by M. So it is 0.704 - 0.0089 = 0.695 cc/cubic meter.

The upper end of the interval is the mean added to M. So it is 0.704 + 0.0089 = 0.713 cc/cubic meter.

So

99% confidence interval: (0.695 cc/cubic meter, 0.713 cc/cubic meter).