Question

The grade averages for 10 randomly selected junior college students are listed below. Assume the grade averages are normally distributed. Find a 98% confidence interval for the true mean. 78 80 83 48 76 72 94 75 91 65

Answer 1

Solution :

Given that,

78 80 83 48 76 72 94 75 91 65

n = 10

= = 762 / 10 = 76.2

s =  (x - )² / n - 1 = 14.7

Point estimate = sample mean = = 76.2

sample standard deviation = s = 13.1

Degrees of freedom = df = n - 1 = 10 - 1 = 9

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t _/2,df = t_0.01,9 = 2.821

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

= 2.821 * (13.1 / 8)

= 11.7

The 98% confidence interval estimate of the population mean is,

- E < < + E

76.2 11.7 < < 76.2 + 11.7

64.5 < < 87.9