Question

# A survey conducted five years ago by the health center at a university showed that 18%...

A survey conducted five years ago by the health center at a university showed that 18% of the students smoked at the time. This year a new survey was conducted on a random sample of 200 students from this university, and it was found that 50 of them smoke. We want to find if these data provide convincing evidence to suggest that the percentage of students who smoke has changed over the last five years. What are the test statistic (Z) and p-value of the test?

Select one:

a. Z=2.58, p-value=0.0049

b. Z=2.29, p-value=0.0220

c. Z=2.29, p-value=0.0110

d. Z=2.58, p-value=0.0098

answer is d. I have no idea why though

A survey conducted five years ago by the health center at a university showed that 18% of the students smoked at the time. This year a new survey was conducted on a random sample of 200 students from this university, and it was found that 50 of them smoke. We want to find if these data provide convincing evidence to suggest that the percentage of students who smoke has changed over the last five years. The p-value of the test is smaller than the significance level, 0.05.

1) Find the conclusion of the test.

2) Do you expect that the 95% confidence interval for the sample proportion will contain 18%?

Select one:

a. 1) Fail to reject the null. 2) The confidence interval will contain 18%.

b. 1) Reject the null. 2) The confidence interval will contain 18%.

c. 1) Fail to reject the null. 2) The confidence interval will not contain 18%.

d. 1) Reject the null. 2) The confidence interval will not contain 18%.

The Hypotheses are:

The test statistic is calculated as

and P-value corresponding to the Z score calculated using Z table shown below as

P-value=0.0098

1) Conclusion:

Since the P-value is <0.05 hence we reject the null hypothesis and conclude there is enough evidence to support the claim.

2) Confidence interval is calculated as

Hence CI= {0.19, 0.31]

So, the conclusion:

d. 1) Reject the null. 2) The confidence interval will not contain 18%.

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