Question

1.) A survey​ asks, "If the husband in a family wants​ children, but the wife decides that she does not want any​ children, is it all right for the wife to refuse to have​ children?" Of 746 subjects, 568 said yes.

A.) Find a 99% confidence interval for the population proportion who would say yes (Round to four decimal places)

B.) Can you conclude that the population proportion exceeds 75%?

1.) No, we cannot conclude that the population proportion exceeds 75% because 75% is below the lowest believable value of the confidence interval.

2.) Yes, we can conclude that the population proportion exceeds 75% because 75% is below the lowest believable value of the confidence interval.

3.) No, we cannot conclude that the population proportion exceeds 75% because 75% is above the lowest believable value of the confidence interval.

4.) Yes, we can conclude that the population proportion exceeds 75% because 75% is above the lowest believable value of the confidence interval.

C.) Without doing any calculation, explain whether the interval in (a) would be wider or narrower than a 95% confidence interval for the population proportion who would say yes.

1.) The 99% confidence interval would be narrower than a 95% confidence interval.

2.) The 99% confidence interval would be wider than a 95% confidence interval.

Answer 1

A) The confidence interval is given as

Here, we have p = 568/746 = 0.76; N = 746

From a standard normal table, we can see that this value corresponds to a z-value of 2.57

Hence the confidence interval is given as

So the 99% confidence interval is [0.72, 0.80]

B) No, we can not conclude that the population proportion exceeds 75% because 75% is above the lowest believable value of the confidence interval. Option 3) is correct.

C) The 99% confidence interval would be wider than a 95% confidence interval. Option 2) is correct.