1. A manufacturer of PVC irrigation pipes claims that the mean bursting pressure of their pipes is 425 psi (pounds per square inch). A consumer watchdog group fears that the bursting pressure may be lower and tests a random sample of 10 such pipes. The bursting pressures are listed below.

1. Define the null and alternative hypothesis for a significance test to help decide if the bursting pressure is lower than claimed by the manufacturer.

H₀: m = 425, H_a: m < 425, T-test.
2. Carry out the test described in part (a) and report the p-value of this test.

p = 0.0422
3. Explain in common language what this p-value tells you.

If the mean bursting pressure is really 425, then a sample like ours (or worse) happens only 4 times in 100, so it unlikely that m = 425.
4. Do you think that the watchdog group has a legitimate fear?

Yes, the results are significant at the 5% level.
5. Based on the sample above, give a 95% confidence interval for the mean bursting pressure of this type of PVC pipe. (386.65, 427.95)

 
2. An agricultural researcher reasons as follows: A heavy application of potassium fertilizer to grasslands in the spring seems to cause lush early growth but depletes the potassium before the growing season ends. Spreading the same amount of potassium over the growing season might increase yields. He therefore compares two treatments:
treatment 1: 100 lbs per acre in the spring;
treatment 2: 50, 25 and 25 pounds per acre applied in spring, early summer and late summer.
The experiment is continued over several years because grass yields vary greatly from year to year. The yields (in pounds of dry matter per acre) are known to vary roughly with a normal distribution (over all years). The data observed in the experiment are given below.
 

Treatment	year 1	year 2	year 3	year 4	year 5
1	3902	4281	5135	5350	5746
2	3970	4271	5440	5490	6028

1. Why is a one sample test appropriate here? The things in the sample are the years, and the years are the same for both treatments. Also, the data is clearly in a matched pairs format.
2. Do the data give good evidence that treatment 2 leads to higher average yields? Fully describe the test you use to answer this question. A right tailed T-test on treatment 2 - treatment 1 of m = 0 against m > 0 has a p-value of 0.03042, so the data give strong evidence that treatment 2 is better.
3. Give a 98% confidence interval for the mean increase in yield due to spreading the potassium applications over the growing season. (-70.34, 384.34)
3. The store manager of the local grocery store claims that customer complaints vary normally from week to week with a mean of 13 complaints per week and a standard deviation of 3 complaints per week. After hiring several new checkout clerks, the manager is worried that the average number of complaints per week has increased. The manager decides to test this hypothesis at the 5% level by selecting a random sample of 10 weeks and calculating  = the average number of complaints for these 10 weeks. If the average number of complaints is too high (significant at the 5% level), the manager will fire the new checkers.
1. Define the null and alternative hypothesis and sketch a (well labeled) normal density curve that illustrates the rejection interval (the values of  that will cause the manager to fire the checkers).

H₀: m = 13, H_a: m > 13, s = 3.> 14.56 is the rejection region. My figure is not well labeled.
2. Several possible results from the test above are stated below. In each case, decide if the checkers are fired or not fired.
1. The p-value of the test is 0.09. Not fired.
2. The p-value of the test is 0.02. Fired.
3. The test yields  = 14.56. On the border, too close to call.
4. The test yields  = 13.29. Not fired.
5. The test yields  = 15.71. Fired.
3. Write down the acceptance interval (the values of  that allow the newly hired checkers to continue working).

Accept if  < 14.56.
4. Describe in common words what a type I error means in this case, and what a type II error means in this case.

Type I error: The manager fires the checkers even though the mean number of weekly complaints has not increased. Type II error: The manager does not fire the checkers, but the mean number of weekly complaints has risen.
5. What is the probability of a type II error if m = 15.5? What is the power of this test against the alternative m = 15.5? About 86% is the power.
4. Monthly salaries (in dollars) of recent graduates with a statistics course are known to vary normally with a standard deviation of $323 per month. You would like to estimate the mean monthly salary of all such graduates with a 90% confidence interval and a margin of error of $150. How large should your sample be? n = 13.

 
5. Which of the following data sets are safe for using T-procedures? For the ones that are not safe, explain why they are not safe. For the ones that are safe, calculate , S_X, and a 95% confidence interval for the population mean.
1. A random sample of size 142 is selected from a population that is significantly skewed to the right.

Safe since n is large, but not enough info to do the other things asked for.
2. The data are: 80,22,17,131,-19,3,23,-1,20,-51,-3

Not safe, the data has large gaps.
3. The data are: 11,18,1,4,6,19,6,3,13,1,-1

The data is not very symmetric, but there are no serious outliers nor large gaps, so T procedures are OK. The sample mean is 7.364, Sx = 6.92 and the confidence interval is (2.72, 12.01).
 
6. Two experimental diets (diet 1 and diet 2) designed to add weight to malnourished 3rd world children are fed to independent and random samples of such children. The results in weight gains are summarized in the table below. Do the data support the claim that diet 2 adds significantly more weight than diet 1? Fully describe and carry out the test you use to make this decision.

Diet 1	Diet 2
1 = 5.80 pounds	2 = 7.27 pounds
n1 = 8	n2 = 9
S1 = 1.613 pounds	S2 = 0.99 pounds

Look at  X1 = diet 1 and X2 = diet 2 in a 2-sample T-test. Null hypothesis is m1 = m2, alternative is m1 < m2. The p-value is 0.023 and the t-statistic is t = -2.23. This is quite strong evidence against H0, or strong evidence that children add more weight with diet 2 than with diet 1.