## Question 563:

1

We're told the population standard deviation is 3.7 hours so we can use either a 1-sample Z test or 1-sample t test.. It is unusual to know the population standard deviation so the t-test is most commonly used. Since the sample size is large, we'll have very similar results regardless of which test we use and I believe the questions is phrased so we are to understand the standard deviation of 3.7 hours is the population standard deviation. So I'll proceed with the 1-sample Z-test here

We want to see if the sample mean of 20.8 hours is sufficient evidence for us to reject the null hypothesis, which is the mean time to install is less than or equal to 20 hours.

1. The Null hypothesis is that the mean time to install siding is less than or equal to 20 hours. The alternative hypothesis is that the mean time is longer than 20 hours.
2. We will reject the null hypothesis if the test statistic is greater than the critical value. The critical value for an alpha of .05 can be found using the percentile to z-score calculator or z-table and its 1.6452 (1-sided).

c. The test statistic:

1. The test statistic is derived by subtracting the test mean from the observed mean (20.8-20) = .8 and dividing this result by the standard error of the mean.
2. The standard error of the mean is the standard deviation divided by the square root of the sample size. So the SEM is 3.7/SQRT(40) = .585.
3. So the z-statistic is .8/.585 = 1.3675.

d. Since the test statistic 1.3675 is less than the critical value of 1.6452 we fail to reject the null hypothesis. We can also look up 1.3675 using a table of normal values or the z-score to percentile calculator and choose one-sided area. It provides us with 8.5734% which is a p-value of .0857. Since this is higher than the .05 cut-off we would also NOT reject the null hypothesis. In other words, there is not sufficient evidence to make the claim that it takes longer than 20 hours to side a house.