A random sample of n_1 = 14 winter days in Denver gave a sample mean

Emily-Jane Bray

Emily-Jane Bray

Answered question

2021-05-05


A random sample of n1=14 winter days in Denver gave a sample mean pollution index x1=43.
Previous studies show that σ1=19.
For Englewood (a suburb of Denver), a random sample of n2=12 winter days gave a sample mean pollution index of x2=37.
Previous studies show that σ2=13.
Assume the pollution index is normally distributed in both Englewood and Denver.
(a) State the null and alternate hypotheses.
H0:μ1=μ2.μ1>μ2
H0:μ1<μ2.μ1=μ2
H0:μ1=μ2.μ1<μ2
H0:μ1=μ2.μ1μ2
(b) What sampling distribution will you use? What assumptions are you making?

The Student's t. We assume that both population distributions are approximately normal with known standard deviations.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference μ1μ2. Round your answer to two decimal places.)

(d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(e) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α?
At the α=0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the α=0.01 level, we reject the null hypothesis and conclude the data are statistically significant.
At the α=0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the α=0.01 level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. (g) Find a 99% confidence interval for
μ1μ2.
(Round your answers to two decimal places.)
lower limit
upper limit
(h) Explain the meaning of the confidence interval in the context of the problem.
Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.

Answer & Explanation

Bella

Bella

Skilled2021-05-06Added 81 answers

Step 1
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Step 2

a) The null and alternative hypotheses are shown below: Null hypothesis:
H0:μ1=μ2 Alternative hypothesis:
H1:μ1μ2

b) Here, the assumptions are both population distributions are approximately normal with known standard deviations. Therefore, the sampling distribution will use here is the standard normal. Thus, the correct option is the standard normal. We assume that both population distributions are approximately normal with known standard deviations.

c) The test statistic is obtained as follows:
z=x1x2σ12n1+σ22n2
z=433719214+13212=0.95

Thous, the value of the test statistics is 0.95.

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