A research firm needs to estimate within 3% the proportion of junior executives leaving large manufacturing companies within three years. A 0.95 degree of confidence is to be used. Several years ago, a study revealed that 35% of junior executives left their company within three years. To update this study, how many junior executives should be surveyed

972 junior executives should be surveyed.

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 .

The margin of error is of:

35% of junior executives left their company within three years.

This means that

0.95 = 95% confidence level

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

To update this study, how many junior executives should be surveyed?

Within 3% of the proportion, which means that this is n for which M = 0.03. So

Rounding up:

972 junior executives should be surveyed.

7 ft 6 in means 7 1/2, because 6 in is 1/2 of a foot. 7 ft 8 in means 7 2/3 because 8 in is 2/3 of a foot. 23/3 = 7 2/3. 15/2 = 7 1/2.

23/3   *   15/2   = 57 1/2

step-by-step explanation: