In: Statistics and Probability
10. The grades of students are normal distributed. In a class of 10 students the average grade on a quiz is 16.35, with a standard deviation of 4.15. ( 3 marks ) a) Find the 90% confidence interval for the population mean grade. b) If you wanted a wider confidence interval, would you increase or decrease the confidence level?
Given that,
= 16.35
s =4.15
n = 10
Degrees of freedom = df = n - 1 = 10- 1 = 9
At 90% confidence level the t is ,
= 1 - 90% = 1 - 0.90 = 0.1
/ 2 = 0.1 / 2 = 0.05
t /2,df = t0.05,9 = 1.833 ( using student t table)
Margin of error = E = t/2,df * (s /n)
= 1.833 * ( 4.15/ 10) = 2.4055
The 90% confidence interval estimate of the population mean is,
- E < < + E
16.35 - 2.4055< <16.35 + 2.4055
13.9445 < < 18.7555
( 13.9445 , 18.7555)
(B) increase