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

Step-by-step 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 :-

$\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}$

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:

$95\pm (2.756)\dfrac{6.6}{\sqrt{30}}\\\\=95\pm (2.756)\dfrac{6.6}{5.47722557}\\\\=95\pm (2.756)(1.205)\\\\=95\pm 3.32\\\\= (95-3.32, 95+3.32)\\\\=(91.68,\ 98.32)$

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

So correct option is c) .