Question

A computer manufacturer estimates that its line of minicomputers has, on average, 8.6 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 1.1 days. Assuming that number of downtime days is normally distributed, test to determine whether these minicomputers actually average 8.6 days of downtime in the entire population. Let α = .01.

Answer 1

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H₀: The minicomputers have 8.6 days of downtime per year on an average.

Alternative hypothesis: H_a: The minicomputers don’t have 8.6 days of downtime per year on an average.

H₀: µ = 8.6 versus H_a: µ ≠ 8.6

This is a two tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 8.6

Xbar = 5.2

S = 1.1

n = 7

df = n – 1 = 6

α = 0.01

Critical value = - 3.7074 and 3.7074

(by using t-table or excel)

t = (Xbar - µ)/[S/sqrt(n)]

t = (5.2 - 8.6)/[1.1/sqrt(7)]

t = -8.1778

P-value = 0.0002

(by using t-table)

P-value < α = 0.01

So, we reject the null hypothesis

There is not sufficient evidence to conclude that the minicomputers have 8.6 days of downtime per year on an average.