Answer: b. 8.56 to 21.44
Step-by-step explanation:
Let [tex]\mu[/tex] 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 : [tex]\overline{x}=15[/tex]
Sample standard deviation : [tex]s=9[/tex]
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
[tex]t_{\alpha/2, 9}=t_{0.025,9}=2.2622[/tex]
Confidence interval for [tex]\mu[/tex] is given by :-
[tex]\overline{x}\pm t*\dfrac{s}{\sqrt{n}}[/tex]
[tex]=15\pm (2.2622)\dfrac{9}{\sqrt{10}}\\\\\approx 15\pm 6.44=(15-6.44,\ 15+6.44)\\\\=(8.56,\ 21.44)[/tex]
Hence, the 95% confidence interval for the true mean number of pushups that can be done is 8.56 to 21.44.
