Question

In a study of monthly salary distribution of residents in Paris conducted in year 2015, it was found that the salaries had an average of €2200 (EURO) and a standard deviation of €550. One years later (in 2016), it was suspected that the average salary had increased. A hypothesis test was conducted at a significance level of 5% to test the suspicion. A random sample of size 64 was chosen with mean of €2385 and standard deviation of €650.

Question 1: Set up the alternative hypothesis.

Select one:

x⎯⎯⎯>2200x¯>2200

μ>2200μ>2200

μ>2385μ>2385

x⎯⎯⎯>2385

Question 2: Assume that the population standard deviation had not changed from 2015. Find the test statistic.

Question 3: Based on the assumption in Question 2 above, find the rejection region.

Select one:

z>1.645z>1.645

z<−1.645z<−1.645

z>1.96z>1.96

z<−1.96z<−1.96

z>2.33z>2.33

z<−2.33

Question 4: Based on the assumption in Question 2, find the p-value.

Question 5: What is the conclusion?

Select one:

There is not enough evidence to infer that the average salary had increased at 5% significance level.

There is enough evidence to infer that the average salary had increased at 5% significance level.

There is enough evidence to infer that the average salary had increased at 10% significance level.

There is not enough evidence to infer that the average salary had increased at 10% significance level.

There is enough evidence to infer that the average salary had increased at 2.5% significance level.

Question 6: The assumption made in Question 1 was certainly not reasonable. Find the test statistic again, assuming that the population standard deviation had changed from 2015. Also assume that the population was normal.

Question 7: Find the rejection region again based on the new assumptions made in Question 6.

t>1.6706t>1.6706

t>1.6686t>1.6686

t>1.9971t>1.9971

t>2.0003

In another study conducted in 2016, the average of monthly salaries of 45 randomly chosen residents in Bordeaux was found to be €3200; the sample standard deviation of the chosen salaries was €600. The suspicion was that the average salary of residents in Paris is lower than the average salary of residents in Bordeaux. The research team was to conduct a F-test to test the suspicion. But they had no information on the two population variances. So they first conducted an F-test at a significance level of 5% to determine if the two population variances are equal or not, assuming that both populations were normal. Set 1 = Paris; 2 = Bordeaux.

Question 8: Set up the alternative hypothesis for the F-test.

Select one:

s21=s22s12=s22

s21≠s22s12≠s22

σ21=σ22σ12=σ22

σ21≠σ22

Question 9: In the F-test, what is the test statistic?

Question 10: Find the rejection region in the F-test.

Question 11: What is the conclusion?

Select one:

There is enough evidence to infer that H1H1 is true.

There is not enough evidence to infer that H1H1 is true

Question 12: The analyst then conducted an appropriate F-test at a significance level of 5% to test the suspicion. Set up the alternative hypotheses for the F-test.

Select one:

x⎯⎯⎯1−x⎯⎯⎯2>0x¯1−x¯2>0

x⎯⎯⎯1−x⎯⎯⎯2<0x¯1−x¯2<0

μ1−μ2>0μ1−μ2>0

μ1−μ2<0

Question 13: Find the test statistic in the F-test.

Question 14: Find the rejection region in the F-test.

Question 15: What is the conclusion?

Select one:

There is enough evidence to infer that μ1μ1 is lower than μ2μ2 at 5% significance level.

There is not enough evidence to infer that μ1μ1 is lower than μ2μ2 at 5% significance level.

There is not enough evidence to infer that x⎯⎯⎯1x¯1 is lower than x⎯⎯⎯2x¯2 at 5% significance level.

There is enough evidence to infer that x⎯⎯⎯1x¯1 is lower than x⎯⎯⎯2x¯2 at 5% significance level.

There is enough evidence to infer that x⎯⎯⎯1x¯1 is lower than x⎯⎯⎯2x¯2 at 2.5% significance level.

Answer 1

SOLUTION 1: The alternative hypothesis.μ>2200

SOLTION 2: TEST STAISTIC Z= XBAR-MEAN/\sqrt(n)

Z=2385-2200/550/sqrt(64)

Z= 185/550/8

Z= 185/68.75

Z=2.691

SOLUTION 3] The rejection region.is Z>1.645.

SOLUTION 4] The P-Value is .003562.The result is significant because p < .05.

SOLUTION 5] There is enough evidence to infer that the average salary had increased at 5% significance level..

SOLUTION 6] TEST STATISTIC t= xbar-mean/s/sqrt(n)

t= 2385-2200/650/8

t= 185/650/8

t=2.277

SOLUTION 7] REJECTION REGION: t>1.6706

NOTE: AS PER THE Q&A GUIDELINES I HAVE DONE THE FIRST PART PLEASE REPOST THE REST.THANK YOU.