Exam 6: The Standard Deviation As a Ruler and the Normal Model

Haircuts You need to find a new hair stylist and know that there are two terrific salons in your area, Hair by Charles and Curl Up & Dye. You want a really good haircut, but you do not want to pay too much for the cut. A random sample of costs for 10 different stylists was taken at each salon (each salon employs over 100 stylists). a. Indicate what inference procedure you would use to see if there is a significant difference in the costs for haircuts at each salon. Check the appropriate assumptions and conditions and indicate whether you could or could not proceed. (Do not do the actual test.) b. A friend tells you that he has heard that Curl Up & Dye is the more expensive salon. i. Write hypotheses for your friend's claim. ii. The following are computer outputs. Which output is the correct one to use for this test? Explain. $\text { Output A: }$ $\text { Two-sample T' for Hair by Charles vs Curl Up \& Dye }$ N Mean StDev SE Mean Hair by Charles 10 22.10 6.33 2.0 Curl Up E Dye 10 26.00 4.81 1.5 Difference $=$ mu (Hair by Charles) - mu (Curl Up & Dye) Bstimate for difference: -3.90000 958 CI for difference: ( $-9.22983,1.42983)$ T-Test of difference $=0$ (vs not $=$ ): T-Value $=-1.55 \quad$ P-Value $=0.140$ $\mathrm{DF}=16$ $\text { Output A: }$ $\text { Paired T for Hair by Charlea - Curl Up \& Dye }$ N Mean St.Dev SB Mean Hair by Charles 10 22.1000 6.3325 2.0025 Curl Up \& Dye 10 26.0000 4.8074 1.5202 Difference 10 -3.90000 7.37036 2.33071 958 CI for mean difference: $(-9.17244,1.37244)$ T-Test of mean difference $=0(\mathrm{vs}$ not $=0): \mathrm{T}$ -Value $=-1.67$ P-Value $=$ 0.129 iii. Use the appropriate computer output to make a conclusion about the hypothesis test based on the data. Make sure to state your conclusion in context.

A researcher found that a 98% confidence interval for the mean hours per week spent studying by college students was (13, 17). Which is true? I. There is a 98% chance that the mean hours per week spent studying by college students is between 13 and 17 hours. II. 98% of college students study between 13 and 17 hours a week. III. Students average between 13 and 17 hours per week studying on 98% of the weeks.

Create and interpret a 95% confidence interval.

Absorption rates into the body are important considerations when manufacturing a generic version of a brand-name drug. A pharmacist read that the absorption rate into the body of a new generic drug (G) is the same as its brand-name counterpart (B). She has a researcher friend of hers run a small experiment to test H₀: µG - µB = 0 against the alternative HA: µG - µB ×0. Which of the following would be a Type I error?

Which statement correctly compares t-distributions to the Normal distribution? I. t distributions are also mound shaped and symmetric. II. t distributions are more spread out than the normal distribution. III. As degrees of freedom increase, the variance of t distributions becomes larger.

The two samples whose statistics are given in the table are thought to come from populations with equal variances. What is the pooled estimate of the population standard deviation? Mean SD 12 45 6 16 41 8

Two agronomists analyzed the same data, testing the same null hypothesis about the proportion of tomato plants suffering from blight. One rejected the hypothesis but the other did not. Assuming neither made a mistake in calculations, which of these possible explanations could account for this apparent discrepancy? I. One agronomist wrote a one-tailed alternative hypothesis, but the other used 2 tails. II. They wrote identical hypotheses, but the one who rejected the null used a higher a-level. III. They wrote identical hypotheses, but the one who rejected the null used a lower a-level.

A random sample of 120 college seniors found that 30% of them had been offered jobs. What is the standard error of the sample proportion?

Carry out the appropriate test and state your conclusion in context.

Which of the statements correctly compares t-distributions to the normal distribution? I. t distributions are also mound shaped and symmetric. II. t distributions have less spread than the normal distribution. III. As degrees of freedom increase, the variance of t distributions becomes smaller.

Writing Scores A private high school is contemplating scheduling writing classes by gender. To assist in this decision, data of recent writing scores are collected and analyzed. It is believed that measurements of writing scores are normally distributed. Write a complete conclusion about writing scores based on the statistical software printout shown below. Two Sample T for MALE vs FEMALE N Mean StDev SEMean 91 50.121 10.305 1.080 109 54.991 8.134 0.779 95\%Cl for -(-7.499,-2.241) = (vs :=-3.6564 P=0.0003DF=169.707

Improving Efficiency A distribution company considers hiring a national training consultant in hopes of improving efficiency in deliveries. The national consultant agrees to work with 30 drivers for one week as part of a trial before the manufacturing company makes a decision about the training program. The training program will be implemented if the average completed number of deliveries increases by more than 12 deliveries per day per driver. The distribution company manager will test a hypothesis using a = 0.02. a. Write appropriate hypotheses (in words and in symbols). b. In this context, which do you consider to be more serious - a Type I or a Type II error? Explain briefly. c. After this trial produced inconclusive results the manager decided to test the training program again with another group of drivers. Describe two changes he could make in the trial to increase the power of the test, and explain the disadvantages of each.

Scrubbers A factory recently installed new pollution control equipment ("scrubbers") on its smokestacks in hopes of reducing air pollution levels at a nearby national park. Randomly timed measurements of sulfate levels (in micrograms per cubic meter) were taken before (Set C1) and after (Set C₂) the installation. We believe that measurements of sulfate levels are normally distributed. Write a complete conclusion about the effectiveness of these scrubbers based on the statistical software printout shown below. SET C1 10.0 8.0 8.0 7.0 6.0 9.0 11.5 8.0 9.5 7.5 5.0 10.0 SET C2 5.0 7.0 1.0 9.0 1.5 5.0 2.5 4.0 9.0 6.0 $\text { Two Sample T for C1 vs C2 }$ Mean StDev SEMean 1 12 8.29 1.83 0.53 2 10 5.00 2.84 0.90 $95 \% \mathrm{Cl} \text { for mu1 - mu2: }(1.20,5.38)$ $\text { T-Test } \mathrm{mu1}=\text { mu2 (vs. not }=\text { ): } \quad T=3.29 \quad \mathrm{P}=0.0037 \quad \mathrm{DF}=20$

Food inspectors need to estimate the level of contaminants in food products packaged at a certain factory. Initial tests were based on a small sample but now the inspectors double the sample size for a follow-up test. The main purpose of the larger sample is to

Every year favorite songs compete to be on a Top 200 list based upon sales and rankings by the experts in the music industry. These songs have many characteristics, such as song length and beats per minute, which vary from category to category in the music industry. A disc jockey wondered if the number of beats per minute in songs classified as dance music were lower than the beats per minute in the songs that are ranked on a Top 200 list from 2001. A random sample of songs from each group was selected and the beats per minute are listed in the chart at the right. Does this sample indicate that songs classified as dance music have lower beats per minute than the songs ranked on a Top 200 list? Dance Songs Top 200 Songs 121119 122120 122121 121118 117122 121121 120119 122123 120119 121118 121118 119120 118120 120124 120123 119 117118 -Create and interpret a 90% confidence interval.

A compensation specialist at a technology company wants to assess the competitiveness of the salaries at his firm. He needs to know the mean salary of the firm's employees. He checks all the employee salaries using company records, and he used his computer to create a 95% confidence interval based on a t-distribution. This procedure was not appropriate. Why?

A random sample of 150 teachers in an inner-city school district found that 72% of them had volunteered time to a local charitable cause within the past 12 months. What is the standard error of the sample proportion?

At one SAT test site students taking the test for a second time volunteered to inhale supplemental oxygen for 10 minutes before the test. In fact, some received oxygen, but others (randomly assigned) were given just normal air. Test results showed that 42 of 66 students who breathed Oxygen improved their SAT scores, compared to only 35 of 63 students who did not get the oxygen. Which procedure should we use to see if there is evidence that breathing extra oxygen can help Test-takers think more clearly?

The average American sees 3.9 movies at the theater each year. A curious student polls 30 friends and family over the course of a week. He finds that his friends have seen an average of 4.5 movies with a standard deviation of 1.2 movies. -Does this sample provide evidence that people are attending the movies more often? Provide a complete significance test to support your answer.

Flight costs Every year Educational Services (ETS) selects readers for the Advanced Placement Exams. Recently the AP* Statistics exam has been graded in Lincoln, Nebraska. One objective of ETS is to achieve equity in grading by inviting teachers to be readers from all parts of the nation. However budgets are a consideration also. The accountants at ETS wonder if the flights from cities west of Lincoln are the same as flight costs from cities east of Lincoln. A random sample of the expense vouchers from last year was reviewed for the cost of airline tickets. Costs (in dollars) are shown in the table. East West 265 257 298 320 340 295 219 288 199 366 398 275 359 430 309 397 105 253 253 366 Indicate what inference procedure you would use to see if there is a significant difference in the costs of airline flights between the west and east coasts to Lincoln, Nebraska, then decide if it is okay to actually perform that inference procedure. (Check the appropriate assumptions and conditions and indicate whether you could or could not proceed. You do not have to do the actual test.)