# In a random sample of 60 refrigerators, the mean repair cost was $150.000 and the standard... ## Question: In a random sample of 60 refrigerators, the mean repair cost was$150.000 and the standard deviation was \$15.50.

Construct a 99% confidence interval for the population mean repair cost.

## Confidence Interval:

When population standard deviation is unknown we use t-distribution to compute the interval for the population mean using the degree of freedom of {eq}(n-1) {/eq}. The higher is the sample size the closer is the critical t to the critical z.

The confidence level {eq}= 0.99 {/eq}

The significance level, {eq}\alpha = 0.01 {/eq}

The sample mean, {eq}\bar{X} = 150 {/eq}

The sample standard deviation, {eq}s = 15.50 {/eq}

The sample size, {eq}n = 60 {/eq}

Degrees of freedom:

{eq}\begin{align*} Df & = n -1\\[1ex] & = 59 \end{align*} {/eq}

Critical value of t using t-distirbution table or the excel =TINV(0.01, 59):

{eq}\begin{align*} t_{critical} & = t_{\alpha/2, df}\\[1ex] & \approx \pm 2.66 \end{align*} {/eq}

{eq}99\% {/eq} confidence interval:

{eq}\begin{align*} \mu & = \bar{X} \pm \frac{t\cdot s}{\sqrt{n}}\\[1ex] & = 150 \pm \frac{2.66\times 15.50}{\sqrt{60}}\\[1ex] & = 150 \pm 5.32 \end{align*} {/eq}

So:

{eq}144.68 < \mu < 155.32 {/eq}