Question

Thirty randomly selected students took the calculus final. If the sample mean was 95 and the standard deviation was 6.6, construct a 99% confidence interval for the mean score of all students.a) 91.69 < μ < 98.31b) 92.95 < μ < 97.05c) 91.68 < μ < 98.32d) 92.03 < μ < 97.97

Answer 1

Answer: c) 91.68 < μ < 98.32

Explanation:

Let μ be the mean score of all students.

Let
`\overline{x}` represents the sample mean score.

As per given , we have

Sample size : n= 30

Sample mean :
`\overline{x}=95`

Sample standard deviation:
`s=6.6`

Significance level
`\alpha=0.01`

Now, Confidence interval of mean when population standard deviation is unknown is given by :-

`\overline{x}\pm t_(\alpha/2, df)(s)/(√(n))`

Degree of freedom = n-1 = 29

Critical two-tailed t-value :
`t_(0.01/2,29)=t_(0.005,29)=2.756` [By students' t distribution table]

Confidence interval of mean will be:

`95\pm (2.756)(6.6)/(√(30))\\\\=95\pm (2.756)(6.6)/(5.47722557)\\\\=95\pm (2.756)(1.205)\\\\=95\pm 3.32\\\\= (95-3.32, 95+3.32)\\\\=(91.68,\ 98.32)`

i.e. a 99% confidence interval for the mean score of all students : 91.68 < μ < 98.32

So correct option is c) .