Question

1.We are interested in conducting a study in order to determine what percentage of voters of a state would vote for the incumbent governor. What is the minimum size sample needed to estimate the population proportion with a margin of error of +/-0.08 at 95% confidence?

A.301

B.150

C.300

D.151

2.We have created a 95% confidence interval for the population mean with the result [10, 15]. What decision will we make if we test H0: m = 16 versus H1: m ¹ 16 at 10% level of significance?

A.Fail to reject H0 in favor of H1

B.Reject H0 in favor of H1

C.We cannot tell what our decision will be from the information given

D.Accept H0 in favor of H1

3.Suppose we want to test H0: m ³ 28 versus H1: m < 28. Which one of the following possible sample results based on a sample of size 36 is most likely to reject H0 in favor of H1?

A.Sample mean = 26, sample standard deviation = 9

B.Sample mean = 24, sample standard deviation = 4

C.Sample mean = 28, sample standard deviation = 2

D.Sample mean = 25, sample standard deviation = 6

4.If a null hypothesis is rejected at the 5% level of significance, it

A.will never be tested at the 1% level

B.will always be rejected at the 1% level

C.may be rejected or not rejected at the 1% level

D.will always be accepted at the 1% level

Answer 1

(1) Since, it is interested in conducting a study in order to determine what percentage of voters of a state would vote for the incumbent governor. Then the minimum size sample needed to estimate the population proportion with a margin of error of +/-0.08 at 95% confidence is calculated as follows:

Given margin of error = E=0.08

The critical value of standard normal test at 95% confidence is Z = 1.96

We assume that percentage of voters of a state would vote(p) and would not vote(q=1-p) for the incumbent governor are equal to 0.5 without bias under the null hypothesis.

Then the minimum sample size

n =(Z/E)2pq = (1.96/0.08)2(0.5)(0.5) = 600.25*0.25 =150.0625=150

=150

Ans:B. 150

(2) It has been created a 95% confidence interval for the population mean with the result  [10, 15]. The decision we will make if we test H0: m = 16 versus H1: m not =16 at 10% level of significance is as follows:

Estimate of mean(sample mean) =(LL+UL)/2 =(10+15)/2 = 12.5

Marginal error = Critical z*SE = (UL-LL)/2=(15-10)/2= 2.5

1.96*SE =2.5 (since Critical value of z at 95% confidence is 1.96)

SE =2.5/1.96 = 1.2755

The calculated standard normal test statistic is

Z = (sample mean-m value under H0)/SE

=(12.5-16)/1.2755 = -3.5/1.2755 = -2.74 = 2.74(without sign)

The critical value of Z at 10% level of significance is 1.28

Since, calculated Z(without sign) 2.74 is more than the critical value1.28 then we reject H0 in favor of H1.

Ans. B. Reject H0 in favor of H1.

(3) Suppose we want to test H0: m = 28 versus H1: m < 28(one tailed test)

Given the sample size n= 36, sqrt(n) =sqrt(36)=6

Suppose the level of significance alfa= 0.05

Test statistic Z = (sample mean-m value under H0)sqrt(n)/ sample sd

For option A, Z = (26-28)6/9 = -1.33 and its p-value = 0.0917

For option B, Z = (24-28)6/4 = -6 and its p-value = 0.00001

For option C, Z = (28-28)6/2 = 0 and its p-value = 0.5

For option D, Z = (25-28)6/6 = -3 and its p-value = 0.00135

Since, for option B, the p-value is very least when compared with other, therefore, the Sample mean = 24, sample standard deviation = 4 is most likely to reject H0 in favor of H1.

Ans. B.Sample mean = 24, sample standard deviation = 4

(4) Given a null hypothesis is rejected at the 5% level of significance.

That is the calculated normal Z(suppose) is greater than the critical value of Z i.e.,1.96 at 5% level of significance.

Since, the critical value of Z at 1% level of significance is 2.58, then the above calculated normal Z is even though it is greater than 1.96 but may or may not be greater than 2.58.

Therefore, null hypothesis which is rejected at the 5% level of significance but may or may not be rejected at the 1% level of significance.

Ans. C.may be rejected or not rejected at the 1% level.