Question

A computer manufacturer believes that its line of minicomputers has, on average, 8.2 days of downtime per year. To test this claim, a researcher contacts seven companies that own one of these computers and is allowed to access company computer records. It is determined that, for the sample, the average number of downtime days is 5.2, with a sample standard deviation of 1.3 days. Test whether these minicomputers actually have, on average, 8.2 days of downtime in the entire population. Let α = .01. (Round your answer to 2 decimal places.)

Answer 1

Answer:

We conclude that minicomputers have, on average, different from 8.2 days of downtime in the entire population.

Step-by-step explanation:

We are given that a computer manufacturer believes that its line of minicomputers has, on average, 8.2 days of downtime per year.

It is determined that, for the sample, the average number of downtime days is 5.2, with a sample standard deviation of 1.3 days.

Let $\mu$ = average days of downtime per year of minicomputers.

So, Null Hypothesis, $H_0$ : $\mu$ = 8.2 days     {means that minicomputers actually have, on average, 8.2 days of downtime in the entire population}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 8.2 days     {means that minicomputers have, on average, different from 8.2 days of downtime in the entire population}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

                            T.S. =  $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample average number of downtime days = 5.2 days

            $\sigma$ = sample standard deviation = 1.3 days

            n = sample of companies = 7

So, the test statistics  =  $\frac{5.2-8.2}{\frac{1.3}{\sqrt{7} } }$  ~ $t_6$

                                     =  -6.106

The value of t test statistics is -6.106.

Now, at 0.01 significance level the t table gives critical values of -3.707 and 3.707 for two-tailed test.

Since our test statistic does not lie within the range of critical values of t, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that minicomputers have, on average, different from 8.2 days of downtime in the entire population.