The manager of Hudson Auto Repair wants to advertise one price for an engine tun
ID: 3251449 • Letter: T
Question
The manager of Hudson Auto Repair wants to advertise one price for an engine tune-up, with parts included. Before he decides the price to advertise, he needs a good estimate of the average cost of tune-up parts. A sample of 10 customer invoices for tune-ups has been taken and the costs of parts, rounded to the nearest dollar, are listed below. 104 74 62 68 97 73 77 65 80 109 a. Indicate the formula you would use to find the point estimate for the mean cost of parts per tune-up for all the turn-ups. b. Construct a 90% confidence interval estimate of the mean cost of parts per tune-up for all of the tune-ups performed at Hudson Auto Repair. A sample of 16 cookies is taken to test the claim that each cookie contains at least 9 chocolate chips. The average number of chocolate chips per cookie in the sample was 7.875 with a standard deviation of 1. Assume the distribution of the population is normal. a. State the null and alternative hypotheses. b. Compute the p-value and test the hypothesis at the 1% level of significance. c. State the conclusion of your test. d. What type of error (type I or II) might you have made?Explanation / Answer
3.
Solution:
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: >= 9
Alternative hypothesis: < 9
Formulate an analysis plan. For this analysis, the significance level is 0.01.
Analyze sample data. Using sample data,
SE = s / sqrt(n) = 1 / sqrt(16) = 0.25
DF = n - 1 = 16 - 1 = 15
t = (x - ) / SE = (7.875 - 9) / 0.25 = -4.5
where s is the standard deviation of the population, x is the sample mean, is the hypothesized population mean, and n is the sample size.
Here is the logic of the analysis: Given the alternative hypothesis ( < 9), we want to know whether the observed sample mean is small enough to cause us to reject the null hypothesis.
We use the t Distribution Calculator to find P(t < -4.5)
The P-Value is 0.000212.
The result is significant at p < 0.01
Interpret results. Since the P-value is smaller than the significance level (0.01), we can reject the null hypothesis.
Conclusion. Reject null hypothesis, there is insufficient evidence to support the claim that each cookie contains atleast 9 chocolate chips.
Type I error is the incorrect rejection of a true null hypothesis (a "false positive"),
while a type II error is incorrectly retaining a false null hypothesis (a "false negative").
We could have made Type II error as the null hypothesis is false and if we would have failed to reject it, then we might have commited this error.
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