HW 5 SOLUTIONS Inference for Two Population Means 1. The Type II Error rate, β = P{failing to reject H 0 H 0 is false}, for a hypothesis test was calculated to be β = 0.07. What is the power = P{rejecting H 0 H 0 is false} for this test? Power = 1 - β = 1-0.07 = 0.93 2. Don is studying species of fish inhabiting a certain river. He fails to reject the null hypothesis that there are an equal amount of species of fish found in the pools and in the riffles of this river. Which of the following statements is true? Don may have made a Type I error. Don definitely made a Type I error. Don may have made a Type II error. Don definitely made a Type II error. Don definitely did not make either a Type I or a Type II error. 3. 7.38. Researchers studied the effect of a houseplant fertilizer on radish sprout growth. They randomly selected some radish seeds to serve as controls, while others were planted in aluminum planters to which fertilizer sticks were added. Other conditions were held constant between the two groups. The following table shows data on the heights of plants (in cm) two weeks after germination. Control Fertilized 3.4 1.6 2.8 1.9 4.4 2.9 1.9 2.7 3.5 2.3 3.6 2.3 2.9 2.8 1.2 1.8 2.7 2.5 2.4 2.7 2.6 2.3 2.2 2.6 3.7 1.6 3.6 1.3 2.7 1.6 1.2 3.0 2.3 3.0 0.9 1.4 2.0 2.3 1.5 1.2 1.8 3.2 2.4 2.6 2.3 2.0 1.7 1.8 2.4 2.6 1.4 1.7 2.5 2.4 1.8 1.5 n 28 28 Y 2.58 2.04 s 0.65 0.72
a. Construct a 90% confidence interval for the difference in mean growth for the control and fertilizer groups. Note, this is the independent samples setting. STAT->TESTS and 2-SampTInt->Stats->Enter->scroll down to enter x1: 2.58 Sx1: 0.65 n1: 28 x2 : 2.04 Sx2:0.72 n2: 28 ->C-Level 0.90-> Pooled NO ->Calculate->yields (0.233, 0.847) b. Interpret the interval you just computed in part (a). We are 90% confident that the true mean height of radish plants two weeks after germination is bigger for radishes with no fertilizer than for radishes grown with fertilizer by at least 0.235 cm and at most 0.851 cm. c. Use the QQplot to assess the normality assumption for this t based interval. Both of the QQplots show a fairly linear pattern with an ever so slight hint of a right skew. Given that these QQplots are not too far from linear and we have 28 observations in each group, we can invoke the CLT. Conclusion: I see no violation of the normality assumption - it is OK to proceed. 4. 7.51. A study was undertaken to compare the respiratory responses of hypnotized and nonhypnotized subjects to certain instructions. The 16 male volunteers were allocated at random to an experimental group to be hypnotized or to a control group. Baseline measurements were taken at the start of the experiment. In analyzing the data, the researchers noticed that the baseline breathing patterns of the two groups were different; this was surprising, since all the subjects had been treated the same up to that time. One explanation proposed for this unexpected difference was that the experimental group were more excited in anticipation of the experience of being hypnotized. The accompanying table presents a summary of the baseline measurements of total ventilation (liters of air per minute per square meter of body area). Experimental Y = 6.169 s = 0.621 5.32 4.5 5.6 4.78 5.74 4.79 6.06 4.86 6.32 5.41 6.34 5.7 6.79 6.08 7.18 6.21 Control Y = 5.291 s = 0.652 a. Construct a 99% confidence interval for the difference in the mean total ventilation. (You may proceed as though the assumptions have been checked and deemed acceptable.) Note, this is the independent samples setting. STAT->TESTS and 2-SampTInt->Stats->Enter->scroll down to enter x1: 6.169 Sx1: 0.621 n1: 8 x2 : 5.291 Sx2:0.652 n2: 8 ->C-Level 0.99-> Pooled NO ->Calculate->yields (-0.07, 1.826)
b. Interpret the interval you just computed in part (a). With 99% confidence, we are unsure whether true mean ventilation is higher for men like these under either condition. If it is higher for men under "control" group settings, it is by as much as 0.070 liters of air per minute per square meter of body area. If it is higher for men under "experimental" group settings, it is by as much as 1.825 liters of air per minute per square meter of body area. 5. 9.38. A volunteer working at an animal shelter conducted a study of the effect of catnip on cats at the shelter. She recorded the number of negative interactions each of 15 cats made in 15 minute periods before and after being given a teaspoon of catnip. The paired measurements were collected on the same day within 30 minutes of one another; the data are given in the accompanying table. Cat Before (Y 1 ) After (Y 2 ) Difference Amelia 0 0 0 Bathsheba 3 6-3 Boris 3 4-1 Frank 0 1-1 Jupiter 0 0 0 Lupine 4 5-1 Madonna 1 3-2 Michelangelo 2 1 1 Oregano 3 5-2 Phantom 5 7-2 Posh 1 0 1 Sawyer 0 1-1 Scary 3 5-2 Slater 0 2-2 Tucker 2 2 0 Mean 1.8 2.8-1 SD 1.66 2.37 1.20 a. Construct a 95% confidence interval for the difference in mean number of negative interactions. STAT->TESTS and TInterval->Enter->Stats->Enter->Scroll down to enter x : -1 Sx: 1.2 n: 15 C-Level:.95 -> Calculate->Enter-> yields (-1.665, -0.336) b. Interpret the interval you just computed in part (a). We are 95% confident that the true mean number of negative interactions is larger for cats that are given catnip by as little as 0.338 interactions or as much as 1.662 interactions.
c. Use the QQplot to assess the normality assumption for the t based interval. Of course, this data cannot come from a normal population, since it is discrete (it is a count). But, we can still check to see if it has a symmetric, bell-shape so that the CLT lets us assume that the distribution of the sample mean is approximately normal. We can see from the stair-step pattern on the QQplot that there are repeated observations (not uncommon in discrete data sets). We like to see at least five categories in discrete data when using the CLT. Taking this into account, the plot looks fairly linear and so the CLT should apply. Conclusion: We proceed with caution, since there are only five categories and 15 observations. 6. 9.34. Thirty-three men with high serum cholesterol, all regular coffee drinkers, participated in a study to see whether abstaining from coffee would affect their cholesterol level. Twenty-five of the men (chosen at random) drank no coffee for 5 weeks, while the remaining eight men drank coffee as usual. A baseline measurement of cholesterol was measured at the beginning of the study. The accompanying table summarizes the change in serum cholesterol levels (in mg/dli) over the five week study. Change from Baseline No Coffee Usual Coffee X -35 26 s 27 56 n 25 8 For the following t distribution based hypothesis tests (t tests), use nondirectional alternatives and let α=0.05. You may proceed as though the assumptions have been checked and deemed acceptable a. The no-coffee group experienced a 35 mg/dli drop in mean cholesterol level. Use a t test to assess whether this difference is significant. Note, this is the paired samples setting. Let N denote no coffee. 2) H 0 : Mean cholesterol does not change in the no coffee condition (μ N,1 = μ N,2 ) H A : Mean cholesterol does change in the no coffee condition (μ N,1 μ N,2 ) STAT->TESTS and T-Test->Enter->Stats->Enter->Scroll down to enter μ0: 0 x : -35 S x : 27 n: 25 μ: μ 0 -> Calculate->Enter 3) t = -6.481 4) P = 1.056 E-6 = 0.000001056 5) Thus, P < α and we reject H 0. 6) There is significant evidence to conclude that true mean cholesterol is different for men after abstaining from coffee for 5 weeks.
b. The usual-coffee group experienced a 26 mg/dli rise in mean cholesterol level. Use a t test to assess whether this difference is significant. Note, this is the paired samples setting. Let U denote usual coffee. 2) H 0 : Mean cholesterol does not change in the usual coffee condition (μ U,1 = μ U,2 ) H A : Mean cholesterol does change in the usual coffee condition (μ U,1 μ U,2 ) STAT->TESTS and T-Test->Enter->Stats->Enter->Scroll down to enter μ 0 : 0 x : 26 S x : 56 n: 8 μ: μ 0 -> Calculate->Enter 3) t = 1.313 4) P = 0.231 5) Thus, P > α and we fail to reject H0. 6) There is not significant evidence to conclude that true mean cholesterol is different for men who drink coffee as usual over a five week period. c. Use a t test to compare the no-coffee change to the usual-coffee change in cholesterol over a five week period. Note, this is the independent sample setting. Let N denote no coffee, U denote usual coffee, and d denote the change from baseline. 2) H 0 : mean change in cholesterol is the same (μ N,d = μ U,d ) H A : mean change in cholesterol is different (μ N,d μ U,d ) STAT->TESTS and 2-SampTTest->Stats->Enter->scroll down to enter x 1 : -35 Sx1: 27 n1: 25 x 2 : 26 Sx2: 56 n2: 8 -> μ 1 : μ 2 ->Pooled NO ->Calculate 3) t = -2.972 4) P = 0.018 5) Thus, P < α and we reject H0. 6) There is significant evidence to conclude that the true mean change in serum cholesterol levels for men at the end of a five week period is different for those who drink coffee as usual and those who abstain from coffee.