Formula to find the confidence interval for population mean :-
[tex]\overline{x}\pm t^*\dfrac{\sigma}{\sqrt{n}}[/tex]
, where [tex]\overline{x}[/tex] = sample mean.
t*= critical z-value
n= sample size.
s= sample standard deviation.
By considering the given question , we have
[tex]\overline{x}= 26[/tex]
[tex]s=6.2[/tex]
n= 50
Degree of freedom : df = 49 [ df=n-1]
Significance level : [tex]\alpha=1-0.95=0.05[/tex]
Using students's t-distribution table, the critical t-value for 95% confidence =[tex]t_{\alpha/2,df}=t_{0.025,49}=2.010[/tex]
Then, 95% confidence interval for the population mean will be :
[tex]26\pm (2.010)\dfrac{6.2}{\sqrt{50}}[/tex]
[tex]=26\pm (2.010)\dfrac{6.2}{7.0710}[/tex]
[tex]=26\pm (2.010)(0.87682)[/tex]
[tex]\approx26\pm1.76[/tex]
[tex]=(26-1.76,\ 26+1.76)=(24.24,\ 27.76)[/tex]
Hence, a 95% confidence interval for the population mean = (24.24, 27.76)
Since 28 is not contained in the above confidence interval , it means it is not reasonable that the population mean is 28 weeks.
