Question

The World Health Organization is planning a study of the average weight gain of Europeans in the last 5 years. Scientists from the WHO plan on taking a SRS of 100 Europeans and calculating a 95% confidence interval of the average weight gain of all Europeans. The WHO changed their minds and now plans on calculating a 90% confidence interval from the sample they will select. Explain to the WHO the impact of changing their confidence level from 95% to 90%. Select all that apply. Their confidence interval would be less likely to capture the sample mean. The probability of selecting a sample which doesn't capture the true value of μ would be 10% rather than 5% if they decide to calculate a 90% confidence interval rather than a 95% confidence interval from the sample they will select. They would increase the margin of error of their confidence interval if they calculated a 90% rather than a 95% confidence interval. They would decrease the margin of error of their confidence interval if they calculated a 90% rather than a 95% confidence interval.

Answer 1

B and D

Explanation:

Options are

A. Their confidence interval would be less likely to capture the sample mean.

B. The probability of selecting a sample which doesn't capture the true value of μ would be 10% rather than 5% if they decide to calculate a 90% confidence interval rather than a 95% confidence interval from the sample they will select.

C. They would increase the margin of error of their confidence interval if they calculated a 90% rather than a 95% confidence interval.

D. They would decrease the margin of error of their confidence interval if they calculated a 90% rather than a 95% confidence interval.

A. Confidence interval is pivoted around mean. So this is an incorrect option.

B. 90% sample values around mean will be included in case of 90% confidence interval and 95% sample values around mean wil be included in case of 95% confidence interval. So this option is correct

C. Margin of error increases with increase in confidence interval as likelihood of a sample value deviating from mean increases. So this is incorrect.

D. same explanation as above