Question

The standard deviation of test scores on a certain achievement test is 11.3. If a randomsample of 81 students had a sample mean score of 74.6, find a 90 percent confidence interval estimatefor the average score of all students.

Answer 1

Answer: (72.535, 76.665)

Step-by-step explanation:

The confidence interval for population mean is given by :-

$\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}$

, where $\sigma$ = population standard deviation.

n= Sample size

$\overline{x}$ = sample mean.

$z^*$ = Critical z-value (two-tailed)

Given : $\sigma=11.3$

n= 81

$\overline{x}=74.6$

The critical values for 90% confidence interval : $z^*=\pm1.645$

Now, the confidence interval for population mean would be :

$74.6\pm (1.645)\dfrac{11.3}{\sqrt{81}}\\\\=74.6\pm(1.645)(\dfrac{11.3}{9})\approx74.6\pm2.065\\\\=(74.6-2.065,\ 74.6+2.065)\\\\=(72.535,\ 76.665)$

Hence, a 90 percent confidence interval estimatefor the average score of all students. : (72.535, 76.665)