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WITCHER [35]
3 years ago
8

A random sample of 250 students at a university finds that these students take a mean of 15.3 credit hours per quarter with a st

andard deviation of 1.6 credit hours. Estimate the mean credit hours taken by a student each quarter using a 95% confidence interval. Round to the nearest thousandth.
Mathematics
1 answer:
SpyIntel [72]3 years ago
5 0

Answer:  15.3\pm0.198

OR

(15.102, 15.498)

Step-by-step explanation:

The formula to find the confidence interval(\mu) is given by :-

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

, where n is the sample size

s = sample standard deviation.

\overline{x}= Sample mean

z_{\alpha/2} = Two tailed z-value for significance level of \alpha .

Given : Confidence level = 95% = 0.95

Significance level = \alpha=1-0.95=0.05

s= 1.6

\overline{x}=15.3

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

By standard normal  distribution table ,

Two tailed z-value for Significance level of 0.05 :

z_{\alpha/2}=z_{0.025}=1.96

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)

Hence, the mean credit hours taken by a student each quarter using a 95% confidence interval. =(15.102, 15.498)

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