Question

In: Statistics and Probability

Patsy Pennypincher is very concerned with the rising cost of education. She collected a sample of...

Patsy Pennypincher is very concerned with the rising cost of education. She collected a sample of 20 institutions (n=20) and found a sample mean cost of $15,500 (mean = 15,500) and a sample standard deviation to be $3000 (s = 3000). Provide Patsy with a 98% confidence interval for this year’s tuition cost.

Please show your work.

Expert Solution

Solution :

Given that,

= 15500

s = 3000

n = 20

Degrees of freedom = df = n - 1 = 20 - 1 = 19

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t_/2,df = t_0.01,19= 2.539

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

= 2.539 * (3000 / $\sqrt$ 20) = 1703.2130

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

- E < < + E

15500 - 1703.2130 < < 15500 + 1703.2130

13796.7870 < < 17203.2130

(13796.7870 , 17203.2130 )

orchestra answered 1 year ago

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