Exam 2 (KEY) July 20, 2009

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1 STAT 2300 Business Statistics/Summer 2009, Section 002 Exam 2 (KEY) July 20, 2009 Name: USU A#: Score: /225 Directions: This exam consists of six (6) questions, assessing material learned within Modules 7.1 through 8.8 of the course textbook. The point total for each question is given within parentheses. You have two (2) hours to complete the exam. To receive full credit, show your work. Clearly indicate your final answer. List of Formulas Confidence Intervals Test Statistics Rejection Point Rules σ x ± z α/2 n z = x µ 0 σ/ z < z α, z > z α, z > z α/2 n s x ± t α/2 t = x µ 0 n s/ t < t α, t > t α, t > t α/2 n ˆp(1 ˆp) ˆp p 0 ˆp ± z α/2 z = z < z α, z > z α, z > z α/2 n p 0 (1 p 0 ) n (n 1)s2 (n 1)s2 (n, χ 2 1)s2 = χ 2 α/2 χ 2 (1 α/2) σ 2 0 χ 2 < χ 2 (1 α), χ2 > χ 2 α, χ 2 > χ 2 α/2 or χ2 < χ 2 (1 α/2) 2 { n = (z + z β )σ zα, if H, z a is one-sided = µ 0 µ a z α/2, if H a is two-sided. 1

2 1. (20) (a) A test of hypotheses, pertaining to the location of the mean (µ) for a quantitative trait, is conducted. The alternative hypothesis for the test is H a : µ < µ 0, where µ 0 is some real number. Suppose we determine the test statistic value to be z = At the 0.05 level of significance, do we reject the null hypothesis? Briefly explain your answer. Solution: We fail to reject H 0. The rejection point rule says to reject H 0 if and only if z < z α = z 0.05 = Since z = 1.65 > 1.645, we fail to reject H 0. The p-value of the test is p-value = P (z 1.65) = Since the p-value is greater than the significance level, α = 0.05, we fail to reject H 0. Fill in the blank for parts (b) to (d) (2pts each). For each, assume the respective probability is calculated under the assumption that the null hypothesis is true. (b) Type I Error, α is the probability of rejecting the null hypothesis, when in fact this hypothesis is true. (c) Power, 1 β is the probability of rejecting the null hypothesis, when in fact this hypothesis is false. (d) Type II Error, β is the probability of failing to reject the null hypothesis, when in fact this hypothesis is false. For parts (e) to (h), circle T for true, or F for false (2pts each). (e) T F The larger the p-value, the more we doubt the null hypothesis. (f) T F Alpha (α) is the probability that the test statistic would assume a value as or more extreme than the observed value of the test. (g) T F If a null hypothesis is rejected at a significance level of 0.05, it will always be rejected at a significance level of (h) T F When the null hypothesis is true, there is no possibility of making a Type I error. 2

3 2. (50) A company manufactures rope whose breaking strengths have a mean of 300 pounds (lb) and a standard deviation of 24 lb. It is believed that by a newly developed process the mean breaking strength can be increased. (a) Setup the appropriate hypotheses to test this claim. Solution: Let µ be the true mean breaking strength for the new ropes. We test H 0 :µ 300 lb H a :µ > 300 lb. (b) If the true mean breaking strength for the new ropes is 7.05 pounds greater than the mean breaking strength for the existing ropes, we want our test (the setup of which you derived in part (a) above) to have 90% power to detect this difference. What sample size (n) is needed? ASSUME: a significance level of 0.05; and that the population standard deviation for the new rope manufacturing process equals that of the existing rope manufacturing process. Solution: Let µ 0 denote the mean breaking strength for the existing ropes. Further, let µ a denote the mean breaking strength for the new ropes. The mean breaking strength for the new ropes (µ a ) is 7.05 pounds greater than the mean breaking strength for the existing ropes (µ 0 ), precisely when µ a µ 0 = 7.05 lb. For 90% power, we assume 10% Type II Error, for which it is z β=0.10 = Our significance level is α = 0.05 for a one-sided test of significance, for which we have z = z α=0.05 = We are told to assume the standard deviation for the new rope manufacturing process equals that of the existing process, so that σ = 24 lb. Thus, 2 n = (z + z β )σ µ 0 µ a ( )(24) = 7.05 = Therefore, we require a sample size of n = 100 ropes. 2 3

4 (c) We draw a random sample of size n from the new rope manufacturing process, where n equals the value you calculated in part (b) above. Suppose we find the mean breaking strength for the sample to be 305 lb. Using the results from this random sample, test your hypotheses (derived in part (a)) at the 0.05 level of significance. Again, assume the population standard deviation for the new rope manufacturing process equals that of the existing rope manufacturing process. What do you conclude? Solution: Since n 30, we get a Central Limit Theorem result. Our test statistic is thus given by z = x µ 0 σ/ n = 24/ 100 = = By part (b) above, we found z 0.05 = Using the rejection point rule, we reject H 0 if and only if z > z Since we reject H 0. The p-value of the test is z = > z 0.05 = 1.645, p-value = P (z 2.083) = These data (the sample of 100 ropes) indicate the true underlying mean breaking strength of ropes under the new manufacturing process, statistically significantly exceeds the value of 300 lb, the mean under the existing manufacturing process (p-value = 0.02). 4

5 3. (30) Body mass index is calculated by dividing a person s weight by the square of his/her height; it is a measure of the extent to which the individual is overweight. For the population of middleaged men who later develop diabetes mellitus, the distribution of baseline body mass indices is approximately normal with an unknown mean µ and unknown standard deviation σ. A sample of 45 men selected from this group has mean x = 25.0 kg/m 2 and standard deviation s = 2.7 kg/m 2. (a) Construct a 95% confidence interval for µ. Solution: First, we are told the distribution of BMI is approximately normal, so that we can base our inference for µ on normal theory. Since σ is unknown, we have to base inference for µ on the t distributions. For 95% confidence, we have 1 α = 0.95, for which α/2 = For n = 45, the corresponding degrees of freedom for our t distribution lookup is df = n 1 = 44. From Table 1.2, we find, t = Thus, a 95% confidence interval for µ is [ ] [ s x ± t α/2 = 25.0 ± (2.0154) 2.7 ] = [24.19 kg/m 2, kg/m 2 ]. n 45 Based on these data, we are 95% confident the true underlying mean BMI for middle-aged men who later develop diabetes mellitus, lies between 24.2 kg/m 2 and 25.8 kg/m 2. (b) At the 0.05 level of significance, test the null hypothesis that the mean baseline body mass index for the population of middle-aged men who later develop diabetes mellitus is equal to 24.0 kg/m 2. What do you conclude? Solution: We are asked to test the null hypothesis, H 0 : µ = 24.0 kg/m 2. Thus, we test the hypotheses H 0 : µ = 24.0 kg/m 2 H a : µ 24.0 kg/m 2. Since: (1) we are testing a two-sided alternative hypothesis at the α = 0.05 level of significance; and (2) the corresponding 95% confidence interval for µ does not contain the null value, µ = 24.0 kg/m 2, we reject H 0. These data indicate, the true underlying mean BMI for middle-aged men who later develop diabetes mellitus, statistically significantly differs from the value 24.0 kg/m 2 (p-value < 0.05). 5

6 4. (50) The manufacturer of the ColorSmart-5000 television set claims 95 percent of its sets last at least five years without needing a single repair. In order to test this claim, a consumer group randomly selects 400 consumers who have owned a ColorSmart-5000 television set for five years. Of these 400 consumers, 316 say their television sets did not need a repair. (a) Find a 99% confidence interval for p, the proportion of all ColorSmart-5000 television sets that have lasted at least five years without needing a single repair. Solution: First, note that ˆp = Next, we find # televisions within the sample lasting at least 5 years Sample size min{nˆp, n(1 ˆp)} = min{316, 84} = 84 > 5, = 316/400 = so that the normal approximation to the sampling distribution for ˆp should be valid. Using Table A.3, we find z = 2.576, for which a 99% confidence interval for p is [ ] [ ] ˆp(1 ˆp) (0.79)(0.21) ˆp ± z = 0.79 ± (2.576) n 400 = [0.79 ± ] = [0.7375, ]. Based on these data, we are 99% confident the true underlying proportion of Colorsmart-5000 television sets lasting at least five years, lies between 73.7% and 84.3%. (b) At the 0.05 level of significance, test the null hypothesis, H 0 : p 0.95, the claim of the manufacturer. What do you conclude? Be sure to setup your hypotheses. Solution: We test the hypotheses H 0 : p 0.95 H a : p < Here, we find min{np 0, n(1 p 0 )} = min{400(0.95), 400(0.05)} = min{380, 20} = 20 > 5, so that, under H 0, the sampling distribution of ˆp is approximately normal with mean, µˆp = p 0 = 0.95 and standard deviation, σˆp, given by σˆp = p0 (1 p 0 ) n = 0.95(0.05) 400 =

7 The test statistic for our test is z = ˆp p 0 p 0 (1 p 0 ) n = (0.95)(0.05) 400 = Using the rejection point rule, we reject H 0 whenever z < z α. It is, = z < = z 0.05, so that we reject H 0. These data indicate, the true underlying proportion of television sets lasting at least 5 years without requiring repair, is statistically significantly less than the claimed value of the manufacturer, 95% (p-value < 0.001). Exact P-Value Calculation: Within the sample of size 400 television, let x represent the number of televisions which do not require a single repair over the initial five year ownership period. Then, x Bin(400, p). For the sample obtained and the alternative stated, the p-value is p-value = P (reject H 0 H 0 is true) = P (x 316 p = 0.95) }{{} observed or more extreme in the direction of the alternative hypothesis 316 ( ) 400 = (0.95) x (1 0.95) 400 x x x=0 =

8 5. (25) Consider a normally distributed population with unknown mean µ and standard deviation σ. We wish to test the hypotheses H 0 :µ = 0.5 mm H a :µ 0.5 mm. A random sample of size two is drawn from this population, resulting in a mean of 0.7 mm and standard deviation of mm. Test these hypotheses at the 0.05 level of significance. What do you conclude? Solution: We are given the set of hypotheses, and we test these hypotheses at the α = 0.05 level of significance. Since the standard deviation is unknown, we base our inference for µ on the t distributions. We have t = x µ 0 s/ n = / 2 = Since the sample size is n = 2, we compare the test statistic value to the t distribution with n 1 = 1 degrees of freedom. From Table 1.2, we find t = We reject H 0 if and only if t > t α/2. It is, = = t < t = , so that we fail to reject H 0. These data indicate, the true underlying mean for this population (µ), is not statistically significantly different from the value 0.5 mm (p-value=0.27). 8

9 6. (50) Consider a normally distributed population with unknown mean µ and standard deviation σ. A random sample of ten measurements from this population yields a mean of 438cm and standard deviation of 2cm. Using the data collected from this random sample: (a) Construct a 95% confidence interval for σ 2. Solution: From Table 1.3 provided, for df = n 1 = 9, we find χ = χ = Therefore, a 95% confidence interval for σ 2 is (n 1)s2 (n 1)s2 (10 1)(2)2 (10 1)(2)2, =, χ 2 α/2 χ 2 (1 α/2) = [1.892cm 2, cm 2 ] Based on these data, we are 95% confident that the true underlying population variance (σ 2 ), lies between 1.89cm 2 and 13.33cm 2. (b) At the 0.05 level of significance, setup the appropriate set of hypotheses and test the null hypothesis, H 0 : σ 2 = 3cm 2, against the two-sided alternative hypothesis. What do you conclude? Solution: At the α = 0.05 level of significance, we test the hypotheses H 0 :σ 2 = 3cm 2 H a :σ 2 3cm 2. Since the 95% confidence interval contains the null value (σ0 2 = 3cm 2 ), we fail to reject the null hypothesis at the 0.05 level of significance. These data indicate, the true underlying variance for this population (σ 2 ), is not statistically significantly different from the value of 3cm 2 (p-value=0.43). 9

10 STAT SUMMER EXAM 2 7/20/ PROBLEM 2 -- SOLUTION (b) 15:16 Wednesday, July 15, 2009 The POWER Procedure One-sample t Test for Fixed Scenario Elements Distribution Normal Method Exact Number of Sides U Null 300 Alpha Standard Deviation 24 Nominal Power 0.9 Computed N Total Actual Power N Total

11 Variable N Lower CL STAT SUMMER EXAM 2 7/20/ PROBLEM 2 -- SOLUTION (c) 15:16 Wednesday, July 15, 2009 Upper CL The TTEST Procedure Statistics Lower CL Std Dev Std Dev Upper CL Std Dev Std Err Minimum Maximum BREAK_STR T-Tests Variable DF t Value Pr > t BREAK_STR

12 Variable N Lower CL STAT SUMMER EXAM 2 7/20/ PROBLEM 3 -- SOLUTION 15:16 Wednesday, July 15, 2009 Upper CL The TTEST Procedure Statistics Lower CL Std Dev Std Dev Upper CL Std Dev Std Err Minimum Maximum BMI T-Tests Variable DF t Value Pr > t BMI

13 STAT SUMMER EXAM 2 7/20/ PROBLEM 4 -- SOLUTION 15:16 Wednesday, July 15, 2009 SUCCESS The FREQ Procedure Frequency Percent Cumulative Frequency Cumulative Percent Binomial Proportion for SUCCESS = 1 Proportion (P) ASE % Lower Conf Limit % Upper Conf Limit Exact Conf Limits 99% Lower Conf Limit % Upper Conf Limit Test of H0: Proportion = 0.95 ASE under H Z One-sided Pr < Z <.0001 Two-sided Pr > Z <.0001 Exact Test One-sided Pr <= P Two-sided = 2 * One-sided 5.511E E-28 Sample Size = 400

14 Variable N Lower CL STAT SUMMER EXAM 2 7/20/ PROBLEM 5 -- SOLUTION 15:16 Wednesday, July 15, 2009 Upper CL The TTEST Procedure Statistics Lower CL Std Dev Std Dev Upper CL Std Dev Std Err Minimum Maximum RESPONSE T-Tests Variable DF t Value Pr > t RESPONSE