The mean time required to repair breakdowns of a certain copying machine is 93 minutes. The company which manufactures the machines claims that breakdowns of its new, improved model are easier to fix. To test this claim, sample of 73 breakdowns of the new model were observed, resulting in a mean repair time of 88.8 minutes and a standard deviation of 26.6 minutes. Use significance level α = 0.05. State clearly what are your null and alternative hypothesis, show your work, and state your conclusion.

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, and we can't concluce that the true mean is less than 93 min at 5% of signficance.

Step-by-step explanation:

Data given and notation

[tex]\bar X=88.8[/tex] represent the sample mean

[tex]s=26.6[/tex] represent the sample standard deviation for the sample

[tex]n=73[/tex] sample size

[tex]\mu_o =93[/tex] represent the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

[tex]p_v[/tex] 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 true mean i lower than 93 min, the system of hypothesis would be:

Null hypothesis:[tex]\mu \geq 93[/tex]

Alternative hypothesis:[tex]\mu < 93[/tex]

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (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:

[tex]t=\frac{88.8-93}{\frac{26.6}{\sqrt{73}}}=-1.349[/tex]

P-value

The first step is calculate the degrees of freedom, on this case:

[tex]df=n-1=73-1=72[/tex]

Since is a one side test the p value would be:

[tex]p_v =P(t_{(72)}<-1.349)=0.0908[/tex]

Conclusion

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, and we can't concluce that the true mean is less than 93 min at 5% of signficance.