A survey of 542 consumers reveals that 301 favor the new design for a product. Construct a 90% confidence interval for the true proportion of all consumers who favor the design.

The 90% confidence interval for the true proportion of all consumers who favor the design is (0.52, 0.59).

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:

90% 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 90% confidence interval for the true proportion of all consumers who favor the design is (0.52, 0.59).

it is x equals 3   plese mark me really need it

step-by-step explanation:

not sure but it may be b, as i said i'm not sure.