Answer:
a) The value t=1.73 < 1.833 fall in the acceptance region, so the effect is not significant. The null hypothesis can not be rejected.
The claim that the new method is better doesn't have enough statistical evidence.
b) P-value=0.06
c) 95% CI

Explanation:
A) We have to test the hypothesis about the difference of the means.
The claim is that the new method gives larger times between repairs.
1) State the null and alternative hypothesis

The significance level is 0.05.
2) The sample mean and standard deviation of the current method is

The sample mean and standard deviation of the new method is

The difference of means is

The standard deviation of the difference of means can be estimated as:

3) Then, we can calculate the t-statistic as:

4) For 9 degrees of freedom and a significance level of 0.05, the critical value for t is t=1.833.
5) The value t=1.73 < 1.833 fall in the acceptance region, so the effect is not significant. The null hypothesis can not be rejected.
The claim that the new method is better doesn't have enough statistical evidence.
B) For a df=10-1=9 and t=1.73, the P-value is

C) In this case, for 9 degrees of freedom and 95% CI, the critical value of t is t=2.262.
The CI can be written as:
