|1.||As part of a quality improvement program, your mail-order company is studying the process of filling customer orders. According to company standards, an order is shipped on time if it is sent within 3 working days of the time it is received. You select a simple random sample (SRS) of 100 of the 5000 orders received in the past month for an audit. The audit reveals that 86 of these orders were shipped on time. Find a 95% confidence interval for the true proportion of the months orders that were shipped on time.|
|2.||Sue, a starting player for a major
college basketball team, made only 38.4% of her free
throws last year. During the summer she worked on
developing a softer shot in the hope of improving her
free-throw accuracy. In the first eight games of this
season Sue made 25 free throws in 40 attempts. Let p
be the probability of making each free throw she shoots
a) State the null hypothesis that Sues free-throw probability has remained the same as last year and the alternative that her work in the summer resulted in a higher probability of success.
b) Calculate the test statistic for testing these hypotheses.
c) Find the P-value. Do you accept or reject the null hypothesis for a = 0.05? At a = 0.01?
d) Give a 90% confidence interval for Sues free-throw success probability this season. Are you convinced that she is now a better free-throw shooter than last season?
e) What assumptions are needed for the validity of the test and confidence interval calculations that you performed?
|3.||An entomologist samples a field for egg masses of a harmful insect by placing a yard-square frame at randomly locations and carefully examining the ground within the frame. A simple random sample of 75 locations selected from a countys pastureland found egg masses in 13 locations. Give a 95% confidence interval for the proportion of all possible locations that are infested.|
|4.||The Gallup Poll asked a sample of 1785 adults, Did you, yourself, happen to attend church of synagogue in the last 7 days. Of the respondents, 750 said Yes. Suppose (it is not, in fact, true) that Gallups sample was random. Give a 99% confidence interval for the proportion of all U.S. adults who attended church or synagogue during the week preceding the poll.|
|5.||A "matched pairs" experiment compares the taste of instant versus fresh-brewed coffee. Each subject tastes two unmarked cups of coffee, one of each type, in random order and states which he or she prefers. Of the 50 subjects who participate in the study, 19 prefer the instant coffee. Let p be the probability that a randomly chosen subject prefers freshly brewed coffee. (In practical terms, p is the proportion of the population who prefer fresh-brewed coffee.) Test the claim that a majority of people prefer the taste of fresh-brewed coffee. State the hypotheses, report the test statistic and the P-value. Is your result significant at the 5% level? What is your practical conclusion?|
|6.||Eleven percent of the products produced by an industrial process over the past several months fail to conform to the specfications. The company modifies the process in an attempt to reduce the rate of nonconformities. In a trial run, the modified process produces 16 nonconforming items out of a total of 300 produced. Do these results demonstrate that the modification is effective? Support your conclusions with a clear statement of your assumptions and the results of your statistical calculations.|
|1.||The estimate is 0.86, SE = 0.0347 so the 95% CI is 0.7920 to 0.9280.|
|2.||a) H0: p = .384 HA:
p > .384 where p is the proportion
of all free throws that Sue makes this season.
b) The test statistic is Z = 3.13.
c) The P-value is 0.0009. Reject the null hypothesis for a = 0.05; also at a = 0.01.
d) .4991 to .7509. There is strong evidence that Sue has improved.
e) We must assume that the shots are equivalent to a random sample from all shots. Also, the sample size must be sufficiently large (which it is).
|3.||The estimate is .173 with SE = .0437, so the 95% CI is 0.0877 to 0.2590.|
|4.||The estimate is .04202 with SE = 0.0117 so the 99% CI is 0.3901 to 0.4503.|
|5.||H0: p = .5 HA: p > .5 where p is the proportion of all people who prefer fresh brewed coffee. The test statistic is 1.70; P-value of 0.0446. The result is statistically significant at the 5% level. The data indicate that fresh brewed coffee is prefered by over half of all coffee drinkers.|
|6.||We test H0: p = .11 HA: p < .11 where p is the proportion of nonconforming items produced by the modified process. We must assume that our trial run produces items that are equivalent to a random sample from all products produced by the modified process. The sample size is sufficiently large. The test statistic is -3.14; the P-value is .0008. Reject H0 at a = .05, also at a =.01 and even at a = .001. This amounts to very strong evidence that HA is the true hypothesis -- that the modification has improved the process.|