MATH 203 Test 3 Review: Hypothesis Testing and Chi-Square Distribution, Exams of Statistics

Download MATH 203 Test 3 Review: Hypothesis Testing and Chi-Square Distribution and more Exams Statistics in PDF only on Docsity! MATH 203 Test 3: Part I – No Calc. Review Problems 1. Below are sample data on SAT scores for college-bound students. The scores are assumed to be normally distributed and the given ’s are based on past tests. Let 1 be the true average SAT score among all college-bound females and let 2 be the true average SAT score among all college-bound males. n x Females 300 1079.6 180 Males 250 1040.5 150 (a) Test H0 : 1 = 2 vs. Ha : 1 > 2 . Use the P -value to explain the conclusion in detail. (b) Test H0 : 1 − 2 = 40 vs Ha : 1 − 2 < 40. Use the P -value to explain the conclusion in detail. (c) Give the numerical formula for the test statistic in Part (b) and state the distribution that it follows. (d) If the scores were not assumed to be normal, or if we had to use S values instead, why could we still use a 2Sample Z–Test? –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– 2. A survey of WKU faculty found that 110 out of 200 (55%) oppose having a Winter term. A survey of staff found that 58 out of 100 (58%) oppose it. Let p1 be the true proportion of faculty who oppose and let p2 be the true proportion of staff who oppose. Use the screens below to see if there is enough evidence to reject the hypothesis that the proportions in opposition are equal. (a) State the null hypothesis and give an appropriate one-sided alternative. (b) Assuming the null hypothesis is true, give the fraction form of the “pooled estimate” of the proportion of Faculty/Staff in opposition to Winter Term. (c) Show how to find the test stat in this case and state the distribution that it follows. (d) For your alternative in Part (a), use the appropriate P -value to explain your conclusion about H0 in detail using = 0.05. (e) To test H0 : p1 – p2 = –0.05, what is the appropriate alternative and what is the test statistic? Explain if you can reject H0 at the 0.05 level of significance. Part II – Calculators Allowed 3. Compute probabilities and bounds of a specific chi-square distribution. The chart of the chi-square scores will be provided. (a) Compute and show a shaded graph the following probabilities. Also label the x - value where the maximum of the distribution curve occurs: (i) P(12 ≤ 2(19) ≤ 25) (ii) P( 2 (17) ≤ 20) (iii) P( 2 (22) ≥ 20) (b) Use the Chi-Square Score Chart to give L and R such that (i) P L ≤ 2(19) ≤ R( ) = 0.90 (ii) P L ≤ 2(17) ≤ R( ) = 0.95 (iii) P L ≤ 2(22) ≤ R( ) = 0.98 –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– 4. Packaged weights of Cheer detergent are normally distributed with mean of 53 ounces and a standard deviation 0.9 ounces. (a) Suppose various random samples of size 40 are collected. Compute P(0.81 ≤ S ≤ 0.99) . (b) Suppose is unknown, but a random sampling of 30 packages yields S = 0.76. Find a 95% confidence interval for the true standard deviation. (c) Using S = 0.86 from a sample of 30 packages, is there evidence at the 10% level of significance, to reject the claim that = 0.9? State initial and alternative hypotheses, give the test-statistic and P -value, and use the P -value to explain your conclusion in detail. What about if S = 1.1 from a sample of size 30? –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– 5. A report claims that consumer preferences for wireless communications are Sprint Verizon T-Mobile Cingular AT&T Other 15% 18% 20% 16% 6% 25% However a random telephone poll of 1200 adults in Kentucky gave the following preferences: Sprint Verizon T-Mobile Cingular AT&T Other 150 210 220 200 70 350 (a) If the report were correct for Kentucky consumers, then what would be the expected numbers of preferences for each category? (b) To test the goodness of fit, compute the Pearson test statistic and the P -value. (c) Is there significant evidence to reject the report’s claim with regard to consumers in KY? Use the P -value to explain your conclusion. ––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––