One-Sided Hypothesis Testing for Mean with Normal Population Distribution, Study notes of Probability and Statistics

Download One-Sided Hypothesis Testing for Mean with Normal Population Distribution and more Study notes Probability and Statistics in PDF only on Docsity! STT Week 10-16-06 Read chapter 7 except “testing population variances” pg. 301. Most important skills are given in pink. Answers in yellow. 1. Each package of a shipment is scored by x =its content weight. We wish to test the hypothesis that mux is at least 128 pounds versus the alternative that mux < 128. The test is to have significance level alpha = 0.01. The test will be based on a with replacement sample of 50 packages from the shipment. a. Is this is a one sided or a two sided test? Ans. This is a one-sided test since the alternative hypothesis H1: mux < 128 lies entirely to one side of the null hypothesis H1: mux at least 128. b. Under what general conditions do we reject the null hypothesis? Ans. Intuiitively, we reject H0 if the sample mean xBAR is too low. c. What is the form of the test statistic for this test? Ans. test statistic = (xBAR – 128) / (sx / root(n)) ~ Z when mux = 128. The test statistic moves up and down with xBAR. So the test will reject H0 if Z is small enough. d. How small must the test statistic be for the test to reject H0? Ans. It must be smaller than the z for which P(Z < z) = alpha = 0.01. Such a z will of course be negative. Using the t-table with column header t.01 and degrees of freedom infinity we find that this z is z = -2.326. So the specific form of the test in this case is to reject H0 if (xBAR – 128) / (sx / root(50)) < -2.326 e. Suppose that a sample of 50 packages is selected with replacement from the shipment and we find xBAR = 122 (this suggests an underweight shipment since 122 is well below 128) with sample sd s = 11. What is the value of the test statistic for this data and what action (reject H0 or do not reject H0) is taken by the test? Ans. The test statistic evaluates to (122-128)/(11/root(50)) = -3.86. Since this is less than –2.326 this test rejects H0: “mux is at least 128.” f. This test might be part of a contract allowing the receiver to literally reject the entire shipment if the null hypothesis is rejected. According to the way the test has been designed, what percentage of shipments meeting the mux = 128 standard will be (falsely) rejected? Ans. Since alpha = 0.01 the test will falsely reject such shipments (with mux = 128) only one percent of the time. g. Evaluate the probability “pSIG” that we’d see an xBAR at least as low as the 122 our sample gave, if truly mu = 128, Ans. P(xBAR < 122 with mux = 128) = P((xBAR – 128)/(sx / root(50)) < (122 – 128)/(sx / root(50) with mux = 128) ~ P(Z < (122 – 128)/(11/root(50))) = P(Z < -3.86) ~ 0.5 – 0.49997 = 0.0003 (we’ve used the closest table entry to z = 3.86) Important: Any test rejects H0 if and only if pSIG < alpha. Here we reject H0 because 0.0003 is indeed less than alpha = 0.01. The same action taken in (e). h. What if the test is fed a sample of 50 from a shipment with mux = 120? How frequently would it reject such a truly underweight shipment? Ans. We can (only) estimate the probability using our sample. P((xBAR – 128) / (sx / root(50) < -2.326 when mux = 122) (remember the test is set up for 128) = P((xBAR – 120) / (sx / root(50) < -2.326 +(128-120)/(sx / root(50) when mux = 122) but we’ve an estimate sx = 11 so we can offer the estimate for the probability above at ~ P(Z < -2.326 + (128-120)/(11/root(50))) = P(Z < 2.82) = 0.5 + P(0 < Z < 2.82) = 0.5 + 0.4976 = 0.9976. So we estimate that around 99.8% of the time this test would (correctly) reject a shipment so underweight as to have mux = 120. We say the power of the test at mu = 122 is 0.9976. 2. Suppose that the sample size in (1) had been n = 5 and the population distribution is normal. As in (1) suppose xBAR = 122 and sx = 11 (for the sample of 5 from a normal population). c. What is the form of the test statistic for this test? Ans. test statistic = (xBAR – 128) / (sx / root(n)) = T with 4 DF when mux = 128. The test statistic moves up and down with xBAR. So the test will reject H0 if t is small enough. d. How small must the test statistic be for the test to reject H0? Ans. It must be smaller than the t for which P(T < t) = alpha = 0.01. Such a t will of course be negative. Using the t-table with column header t.01 and degrees of freedom 4 we find that this t is t = -3.747. So the specific form of the test in this case is to reject H0 if (xBAR – 128) / (sx / root(5)) < -3.747 e. Suppose that a sample of 5 packages is selected from the shipment and we find xBAR = 122 (this suggests an underweight shipment since 122 is well below 128) with sample sd s = 11. What is the value of the test statistic for this data and what action (reject H0 or do not reject H0) is taken by the test? for even very tiny alpha (e.g. z(.005) = 2.576). 7-20 a. One or two sided? ans. I interpret the company’s claim to be H0: mu is at least mu0 = 11.5%. We’ll test vs H1: mu < 11.5. With this setup we are giving the benefit of the doubt to the company. We’ll reject H0 if the evidence is strong enough. b. Use t-entry for 0.05 or 0.025? ans. t0 = -t(alpha/2) = -t(0.05/2) = -t(0.025) = -1.96(for z) or closer to -2.00 if the population is thought to be normal (allowing us to use t) c. Form and evaluation of test statistic. ans. TS = (xBAR-mu0)/(s/root(n)) = (10.8 – 11.5)/(3.4 / root(50)) = -1.46. d. Reject Ho or not? ans. Test rejects Ho if TS < t0 (i.e. z0). We fail to reject since –1.46 is not less than – 1.96. 7-28 Change the language to “that it still controls at least 42%.” a. One or two sided? ans. H0: p at least .42 vs H1: p < .42. One sided test. b. Use t-entry for 0.05 or 0.025? ans. z0 = -z(alpha) = -z(.05) = -1.645. c. Form and evaluation of test statistic. ans. TS = (pHAT – p0)/(root(p0 q0 / n) = (219/550 - .42)/(root(.42 .58 / 550) = -1.037. d. Reject Ho or not? ans. Reject if TS < z0. It is not, so we fail to reject H0. 7-39 a. One or two sided? ans. I read the problem as testing H0: p at least 0.49 (the published rate of midlife crisis) vs H1: p < 0.49. That is, we stay with the published rate unless we have strong evidence that the “program” really reduces the rate of crisis. This is one sided. b. Use t-entry for 0.05 or 0.025? ans. z0 = -z(alpha) = -z(0.05) = -1.645. c. Form and evaluation of test statistic. ans. TS = (pHAT – p0)/(root(p0 q0 / n) = (49/125 - .49)/root(.49 .51 / 125) = -2.19 d. Reject Ho or not? ans. Reject H0 if TS < z0 and indeed it is. So reject H0.