Answer: ![15.3\pm0.198](https://tex.z-dn.net/?f=15.3%5Cpm0.198)
OR
(15.102, 15.498)
Step-by-step explanation:
The formula to find the confidence interval
is given by :-
![\overline{x}\pm z_{\alpha/2, df}\dfrac{s}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%5Cpm%20z_%7B%5Calpha%2F2%2C%20df%7D%5Cdfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D)
, where n is the sample size
= sample standard deviation.
= Sample mean
= Two tailed z-value for significance level of
.
Given : Confidence level = 95% = 0.95
Significance level = ![\alpha=1-0.95=0.05](https://tex.z-dn.net/?f=%5Calpha%3D1-0.95%3D0.05)
![\overline{x}=15.3](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%3D15.3)
sample size : n= 250 , which is extremely large ( than n=30) .
So we assume sample standard deviation is the population standard deviation.
thus , ![\sigma=1.6](https://tex.z-dn.net/?f=%5Csigma%3D1.6)
By standard normal distribution table ,
Two tailed z-value for Significance level of 0.05 :
![z_{\alpha/2}=z_{0.025}=1.96](https://tex.z-dn.net/?f=z_%7B%5Calpha%2F2%7D%3Dz_%7B0.025%7D%3D1.96)
Then, the 95% confidence interval for the mean credit hours taken by a student each quarter :-
![15.3\pm (1.96)\dfrac{1.6}{\sqrt{250}}\\\\ =15.3\pm 0.19833\\\\=\approx15.3\pm0.198\\\\=(15.3-0.198,\ 15.3+0.198)=(15.102,\ 15.498)](https://tex.z-dn.net/?f=15.3%5Cpm%20%281.96%29%5Cdfrac%7B1.6%7D%7B%5Csqrt%7B250%7D%7D%5C%5C%5C%5C%20%3D15.3%5Cpm%200.19833%5C%5C%5C%5C%3D%5Capprox15.3%5Cpm0.198%5C%5C%5C%5C%3D%2815.3-0.198%2C%5C%2015.3%2B0.198%29%3D%2815.102%2C%5C%2015.498%29)
Hence, the mean credit hours taken by a student each quarter using a 95% confidence interval. =(15.102, 15.498)