Question

A statistics teacher started class one day by drawing the names of 10 students out of a hat and asked them to do as many pushups as they could. The 10 randomly selected students averaged 15 pushups per person with a standard deviation of 9 pushups.Suppose the distribution of the population of number of pushups that can be done is approximately normal. The 95% confidence interval for the true mean number of pushups that can be done is
a. 13.02 to 16.98.
b. 8.56 to 21.40.
c. 11.31 to 18.55.
d. 5.75 to 24.25.

Answer 1

Answer:  b. 8.56 to 21.44

Step-by-step explanation:

Let  $\mu$ be the  mean number of pushups that can be done.

As per given , we have

Sample size : n= 10

Degree of freedom = n-1=9

Sample mean : $\overline{x}=15$

Sample standard deviation : $s=9$

Significance level : α=1-0.95=0.05

From t- distribution table ,

Critical two -tailed t-value for α=0.05 and df = 9 is

$t_{\alpha/2, 9}=t_{0.025,9}=2.2622$

Confidence interval for $\mu$ is given by :-

$\overline{x}\pm t*\dfrac{s}{\sqrt{n}}$

$=15\pm (2.2622)\dfrac{9}{\sqrt{10}}\\\\\approx 15\pm 6.44=(15-6.44,\ 15+6.44)\\\\=(8.56,\ 21.44)$

Hence, the  95% confidence interval for the true mean number of pushups that can be done is 8.56 to 21.44.