Question

Refrigerators in a random sample of 60 refrigerators, the mean repair cost was $150.00 and the standard deviation was $15.50. Construct a 99% confidence interval for the population mean repair cost. Interpret the results.

Answer 1

Answer:

The 99% confidence interval is given by ($144.677;$155.323)  

We are confidenct at 99% that the true mean for the reapir cost of refrigerators is between 144.677 and 155.323.

Step-by-step explanation:

1) Notation and definitions  

n=60 represent the sample size  

$\bar X= 150$ represent the sample mean  

$s=15.5$ represent the sample standard deviation  

m represent the margin of error  

Confidence =99% or 0.99

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

2) Calculate the critical value tc  

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$. The degrees of freedom are given by:  

$df=n-1=60-1=59$  

We can find the critical values in excel using the following formulas:  

"=T.INV(0.005,59)" for $t_{\alpha/2}=-2.66$  

"=T.INV(1-0.005,59)" for $t_{1-\alpha/2}=2.66$  

The critical value $tc=\pm 2.66$  

3) Calculate the margin of error (m)  

The margin of error for the sample mean is given by this formula:  

$m=t_c \frac{s}{\sqrt{n}}$  

$m=2.66 \frac{15.5}{\sqrt{60}}=5.323$  

4) Calculate the confidence interval  

The interval for the mean is given by this formula:  

$\bar X \pm t_{c} \frac{s}{\sqrt{n}}$  

And calculating the limits we got:  

$150 - 2.66 \frac{15.5}{\sqrt{60}}=144.677$  

$150 + 2.66 \frac{15.5}{\sqrt{60}}=155.323$  

The 99% confidence interval is given by ($144.677;$155.323)  

We are confidenct at 99% that the true mean for the reapir cost of refrigerators is between 144.677 and 155.323.