Astatistics student would like to estimate the average age of students at her college. she would like to be 85% confident that her estimate includes the actual college population mean. she did a smaller preliminary survey and found that a reasonable estimate of the population standard deviation was 4 1/2 years. how many people must she survey if she wants to be within 1/2 year of the true population mean? round off your answer to the nearest integrer.

A statistics student would like to estimate the average age of students at her college. she would like to be 85% confident that her estimate includes the actual college population mean

It means confidence level is 0.85.

After taking survey the estimate of population standard deviation is found as 4.5 years. The given margin of error is 0.5 years

The sample size is obtained using formula

n=

Where std = standard deviation

alpha = 1 -c = 1-0.85 = 0.15

z(α/2) = z(0.15/2) = z(0.075)

It is the critical value of z such that area above z and below -z is 0.075

Use z score table to find the probability exactly or very close to 0.075 and note the corresponding z score value.

In z score table there is no accurate 0.075 probability value. The close value is 0.0749 which corresponds to z score -1.44

For calculation we take positive z score value.

So we get z(α/2) =1.44

Using all the given values into formula

n = [(1.44 * 4.5) / 0.5]^2

n = 167.96

Rounding it to nearest integer we get n=168

The number of people need to surveyed is n=168

savings account is an account only used for saving money. this acts as an emergency source of funding for an individual. answer a would result in the money in his account to grow as it specifically states that interest is compounded daily at the rate of 2%

step-by-step explanation:

pls refer ro the top for answer