Question

1) The local swim team is considering offering a new semi-private class aimed at entry-level swimmers, but needs at minimum number of swimmers to sign up in order to be cost effective. Last year’s data showed that during 8 swim sessions the average number of entry-level swimmers attending was 15. Suppose the instructor wants to conduct a hypothesis test to test the population mean is less than 15, and we know that the variable is normally distributed. The sample size is 8, population standard deviation (σ) is 3.45 swimmers, and α =.05. The sample average is (x ¯) 12.5 swimmers. For a 5% significance level, what is the interpretation based on the decision?

a. Based on this sample, at 5% significance level, the data does not provide sufficient evidence to support that average number of entry-level swimmers attending was less than 15.

b. The test is inconclusive

c. Based on this sample, at 5% significance level, the data provides sufficient evidence to support that average number of entry-level swimmers attending was less than 15.

d. Based on this sample, at 95% significance level, the data does not provide sufficient evidence to support that average number of entry-level swimmers attending was less than 15.

Based on this sample, at 5% significance level, the data provides sufficient evidence to support that average number of entry-level swimmers attending was 15.

2)

Suppose that we want to test whether the majority of the people thinks remote learning worked out better than they expected. For a sample of 25 people we asked the question "Do you think remote learning worked out better than you expected?", recorded the number of yeses, and calculate the test statistic as 1.89. Which one of the following is the p value for this test and your decision at 5% significance level?

a. p-value=0.0588, bBased on this sample, at 5% significance level, we DO NOT have enough evidence to support that majority thinks remote learning worked out better than they expected.

b. p-value=0.0294, based on this sample, at 5% significance level, we have enough evidence to support that majority thinks remote learning worked out better than they expected.

c. p-value=0.0588, based on this sample, at 5% significance level, we have enough evidence to support that majority thinks remote learning worked out better than they expected.

d. This question cannot be answered without further information.

e. p-value=0.0294, based on this sample, at 5% significance level, we DO NOT have enough evidence to support that majority thinks remote learning worked out better than they expected.

3)

A chain store would like to test the hypothesis that no difference exists between the average smart phone sales before and after they employ a new social network advertising campaign. A random sample of ten stores was selected and their average daily sales before and after the campaign are recorded. Let sample 1 be the sales after the campaign, sample 2 be the sales before the campaign, and d be the difference between sample 1 and sample 2. Which one of the following are the null and alternative hypotheses?

1. H 0 : μ d < 0  vs  H a : μ d > 0
2. H 0 : μ d = 0  vs  H a : μ d > 0
3. H 0 : μ d < 0  vs  H a : μ d = 0
4. H 0 : μ d = 0  vs  H a : μ d ≠ 0

4)

A chain store would like to test the hypothesis that no difference exists between the average smart phone sales before and after they employ a new social network advertising campaign. A random sample of ten stores was selected and their average daily sales before and after the campaign are recorded. Let sample 1 be the sales after the campaign, sample 2 be the sales before the campaign, and d be the difference between sample 1 and sample 2. Based on this sample, we have the following information: d ¯ = 7.8and s d = 16.5. Assuming the population distribution is normal, what is the test statistic?

a. 0.5

b. -3.45

c. 12.4

d. 2.3

e. 1.495

5)

A chain store would like to test the hypothesis that no difference exists between the average smart phone sales before and after they employ a new social network advertising campaign. A random sample of ten stores was selected and their average daily sales before and after the campaign are recorded. Let sample 1 be the sales after the campaign, sample 2 be the sales before the campaign, and d be the difference between sample 1 and sample 2. Based on this sample, we have the following information: d ¯ = 7.8and s d = 16.5. What is the p-value for this test?

a. 0.30 < p-value < 0.40

b. P-value cannot be calculated.

c. 0.10 < p-value < 0.20

d. 0.05 < p-value < 0.10

e. 0.025 < p-value < 0.05

6)

A chain store would like to test the hypothesis that no difference exists between the average smart phone sales before and after they employ a new social network advertising campaign. A random sample of ten stores was selected and their average daily sales before and after the campaign are recorded. Let sample 1 be the sales after the campaign, sample 2 be the sales before the campaign, and d be the difference between sample 1 and sample 2. Based on this sample, we have the following information: d ¯ = 7.8and s d = 16.5. At 10% significance level, what is the decision for this hypothesis test?

a. The test is inconclusive.

b. Based on this sample, at 10% significance level, the data provides sufficient evidence to support that the mean sales before the campaign are lower than the mean sales. after the campaign

c. Based on this sample, at 10% significance level, the data does not provide sufficient evidence to support that the mean sales before the campaign are different than the mean sales after the campaign.

d. Based on this sample, at 10% significance level, the data provides sufficient evidence to support that the mean sales before the campaign are different than the mean sales after the campaign

e. Based on this sample, at 90% significance level, the data provides sufficient evidence to support that the mean sales before the campaign are lower than the mean sales after the campaign.

7)

Suppose now the chain store would like to test the hypothesis that the average smart phone sales increase after they employ a new social network advertising campaign. A random sample of ten stores was selected and their average daily sales before and after the campaign are recorded.Let sample 1 be the sales after the campaign and sample 2 be the sales before the campaign. Based on this sample, we have the following information: d ¯ = 7.8and s d = 16.5. What is the new p-value for this test?

a. 0.01 < p-value < 0.025

b. 0.05 < p-value < 0.10

c. Cannot be determined without further information.

d. It stays the same.

e. 0.5 < p-value < 1

8)

A researcher is interested in testing to determine if the mean price of a casual lunch is different in the city (population 1) than it is in the suburbs(population 2). Which one of the following is her null hypothesis?

1. H 0 : μ 1 − μ 2 < 0
2. H 0 : μ 1 − μ 2 = 0
3. H 0 : μ = 0
4. H 0 : μ 1 − μ 2 > 0
5. H 0 : μ 1 − μ 2 ≠ 0

Answer 1

1)

One-Sample Z

Descriptive Statistics

N	Mean	SE Mean	95% Upper Bound for μ
8	12.50	1.22	14.51

μ: mean of Sample
Known standard deviation = 3.45

Test

Null hypothesis	H₀: μ = 15
Alternative hypothesis	H₁: μ < 15

Z-Value	P-Value
-2.05	0.020

Since p-value is less than alpha 0.05  

" Based on this sample, at 5% significance level, the data provides sufficient evidence to support that average number of entry-level swimmers attending was less than 15. "

option c.

2) H0: p = 0.5

H1: p>0.5

test statistic Z = 1.89

p-value = 1 - P(Z<1.89) = 0.0294

"b. p-value=0.0294, based on this sample, at 5% significance level, we have enough evidence to support that majority thinks remote learning worked out better than they expected."

3) H 0 : μ d = 0  vs  H a : μ d ≠ 0

4) n = 10

d ¯ = 7.8and s d = 16.5

t = d/(sd/sqrt(n)) = 7.8/(16.5/sqrt(10)) = 1.495

5)

df = 10-1 = 9

p-value = 0.169

0.10 < p-value < 0.20

6)  

" c. Based on this sample, at 10% significance level, the data does not provide sufficient evidence to support that the mean sales before the campaign are different than the mean sales after the campaign."
7)

H 0 : μ d = 0  vs  H a : μ d > 0

df = 10-1 = 9

p-value = 0.085

0.05 < p-value < 0.10

8) H 0 : μ 1 − μ 2 = 0