Question

Scores of 50 college students who have taken a statistics test has a mean of 82 and a standard deviation of 5. Construct a 99 % confidence interval for the mean score. Also, report the critical value LaTeX: z_cz c corresponding to a confidence level of c

Answer 1

Answer: (80.1786, 83.8214)

Explanation:

Confidence interval for population mean is given by :-

`\overline{x}\pm z_c(\sigma)/(√(n))` , where
`\overline{x}` = Sample mean , n= sample size ,
`\sigma` = standard deviation,
`z_c` = critical z-value.

Given: n= 50,
`\overline{x}= 82` ,
`\sigma=5`

Critical z-value for 99% confidence level = 2.576

Now, a 99 % confidence interval for the mean:

`82\pm(2.576)(5)/(√(50))\\\\=82\pm(2.576)(0.70710)\\\\=82\pm1.8215\\\\=(82-1.8214,\ 82+1.8214)\\\\=(80.1786,\ 83.8214)`

Hence, required 99% confidence interval = (80.1786, 83.8214)