Question

supervisor records the repair cost for 25 randomly selected dryers. A sample mean of $93.36 and standard deviation of $19.95 are subsequently computed. Determine the 98% confidence interval for the mean repair cost for the dryers. Assume the population is approximately normal. Find the critical value that should be used in constructing the confidence interval.

Answer 1

Answer with explanation:

The confidence interval for population mean (when population standard deviation is unknown) is given by :-

$\overline{x}-t^*\dfrac{s}{\sqrt{n}}< \mu$

, where n= sample size

$\overline{x}$ = Sample mean

s= sample size

t* = Critical value.

Given : n= 25

Degree of freedom : $df=n-1=24$

$\overline{x}= \ 93.36$

$s=\ 19.95$

Significance level for 98% confidence interval : $\alpha=1-0.98=0.02$

Using t-distribution table ,

Two-tailed critical value for 98% confidence interval :

$t^*=t_{\alpha/2,\ df}=t_{0.01,\ 24}=2.4922$

⇒ The critical value that should be used in constructing the confidence interval = 2.4922

Then, the 95% confidence interval would be :-

$93.36-(2.4922)\dfrac{19.95}{\sqrt{25}}< \mu$

$=93.36-9.943878< \mu$

$=93.36-9.943878< \mu$

$=83.416122< \mu$

Hence, the 98% confidence interval for the mean repair cost for the dryers. = $83.4161$