In a survery of 154 households, a Food Marketing Institute found that 106 households spend more than $125 a week on groceries. Please find the 95% confidence interval for the true proportion of the households that spend more than $125 a week on groceries.

The 95% confidence interval for the true proportion of the households that spend more than $125 a week on groceries is (0.6151, 0.7615).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

For this problem, we have that:

95% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

The 95% confidence interval for the true proportion of the households that spend more than $125 a week on groceries is (0.6151, 0.7615).

i think the answer would be c