Of 1016 adults in a certain country responding to apoll, 41% said they always or often read nutrition labels when grocery shopping. Complete parts(a) through(d) below.

a. Find a95% confidence interval for the population proportion of adults in the country who always or often read nutrition labels when grocery shopping. Assume the poll used a simple random sample(SRS). (Infact, it used randomsampling, but a more complex method thanSRS.)

b. What is the width of the95% confidenceinterval?

c. Name a confidence level that would produce an interval wider than the95% confidence interval. Explain why you think this interval would be wider than a95% confidence interval.

d. Construct the interval using the confidence level you proposed in part(c) and find the width of the interval. Is this interval wider than the95% confidenceinterval?

The confidence interval you proposed in part(c) for the population proportion is?

Find the width of this interval.

The width of the confidence interval you proposed in part(c) is?