Question

The amount of shaft wear after a fixed mileage was determined for each of seven randomly selected internal combustion engines, resulting in a mean of 0.0372 in. and a standard deviation of 0.0125 in. a. Assuming that the distribution of shaft wear is normal, test at level 0.05 the hypotheses

Answer 1

`t=(0.0372-0.035)/((0.0125)/(√(7)))=0.466`

`p_v =P(t_(6)>0.466)=0.329`

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 0.035 at 5% of significance.

Explanation:

We assume that the hypothesis that they want to check is :

Null hypothesis:
`\mu \leq 0.035`

Alternative hypothesis:
`\mu >0.035`

Data given and notation

`\bar X=0.0372` represent the sample mean

`s=0.0125` represent the sample standard deviation

`n=7` sample size

`\mu_o =0.035` represent the value that we want to test

`\alpha=0.05` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 0.035, the system of hypothesis are :

Null hypothesis:
`\mu \leq 0.035`

Alternative hypothesis:
`\mu > 0.035`

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

`t=(0.0372-0.035)/((0.0125)/(√(7)))=0.466`

P-value

We can find the degrees of freedom like this:

`df = n-1 = 7-1 =6`

Since is a one-side upper test the p value would given by:

`p_v =P(t_(6)>0.466)=0.329`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 0.035 at 5% of significance.