Question

Suppose you plan to take a survey of math students asked about their overall grades. 22 students were surveyed, and it was found that the average GPA of the 22 sampled students was a 3.2, with a standard deviation of 0.9 points. Suppose you want to re-sample the population to reduce the margin of error. If you plan to calculate a 90% confidence interval for the true mean GPA of math students, how many samples would be necessary to reduce the margin of error to 0.2 points? Round your answer to the nearest whole number, and remember it's always better to sample too many rather than not enough.

Answer 1

Solution :

Given that,

= 3.2

s = 0.9

n = 22

Degrees of freedom = df = n - 1 = 22 - 1 = 21

A ) At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t _/2,df = t_0.05,21 =1.721

Margin of error = E = t_/2,df * (s /n)

= 1.721 * (0.9 / 22)

= 0.33

The 90% confidence interval estimate of the population mean is,

- E < < + E

3.2- 0.33 < < 3.2 + 0.33

2.87 < < 3.53

(2.87, 3.53 )

B ) margin of error = E = 0.2

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z_/2 = Z_0.05 = 2.576

Sample size = n = ((Z_/2 * ) / E)²

= ((1.645 * 0.9 ) / 0.2)²

= 54.79

= 55

Sample size = 55