A computer manufacturer estimates that its line of minicomputers has, on average, 7.2 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 4.4, with a sample standard deviation of 1.1 days. Assuming that number of downtime days is normally distributed, test to determine whether these minicomputers actually average 7.2 days of downtime in the entire population. Let α = .01.

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Answer:

The average of minicomputers is not 7.2 days of downtime per year.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 7.2 days

Sample mean, [tex]\bar{x}[/tex] = 4.4 days

Sample size, n = 7

Alpha, α = 0.01

Sample standard deviation, s = 1.1 days

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 7.2\text{ days of downtime per year}\\H_A: \mu \neq 7.2\text{ days of downtime per year}[/tex]

We use two-tailed t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{4.4 - 7.2}{\frac{1.1}{\sqrt{7}} } = -6.734[/tex]

Now,

[tex]t_{critical} \text{ at 0.01 level of significance, 6 degree of freedom } = \pm 3.707[/tex]

Since,                        

Since, the calculated t-statistic does not lie in the acceptance region, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Conclusion:

Thus, we conclude there is not enough evidence to support the claim that minicomputers actually average 7.2 days of downtime in the entire population.