Question

Records on a fleet of trucks reveal that the average life of a set of spark plugs is normally distributed with a mean of 22,100 miles. The fleet owner purchased 18 sets and found that the sample average life was 23,400 miles; the sample standard deviation was 1,412 miles.

a) To decide if the sample data support the company records that the spark plugs average 22,100 miles, state your decision in terms of the null hypothesis. Use a 0.05 level of significance.

b) What is the critical value for the test using a 0.05 level of significance?

c) What is the test statistic?

d) What is your decision?

Answer 1

a) We want to conduct a hypothesis in order to see if the true mean is 22100 or not, the system of hypothesis would be:

Null hypothesis:
`\mu = 22100`

Alternative hypothesis:
`\mu \\eq 22100`

b) We need to find the degrees of freedom given by:

`df =n-1 = 18-1=17`

And the critical values for this case are:

`t_(\alpha/2)= 2.110`

c)
`t=(23400-22100)/((1412)/(√(18)))=3.906`

d) Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 221100 mi

Explanation:

Information provided

`\bar X=23400` represent the sample mean

`s=1412` represent the sample standard deviation

`n=18` sample size

`\mu_o =22100` represent the value to verify

`\alpha=0.05` represent the significance level

t would represent the statistic (variable of interest)

`p_v` represent the p value

Part a

We want to conduct a hypothesis in order to see if the true mean is 22100 or not, the system of hypothesis would be:

Null hypothesis:
`\mu = 22100`

Alternative hypothesis:
`\mu \\eq 22100`

Part b

We need to find the degrees of freedom given by:

`df =n-1 = 18-1=17`

And the critical values for this case are:

`t_(\alpha/2)= 2.110`

Part c

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the info we got:

`t=(23400-22100)/((1412)/(√(18)))=3.906`

Part d

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 221100 mi