InferentialStats-PracticeProblems-Answer

InferentialStats-PracticeProblems-Answer - STA2023...

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1 STA2023 – Inferential Statistics Practice Problems – Answer Key 1. Use 1-sample t-procedures (chapter 23) because σ is unknown. See the assumptions/conditions on pg. 591-592. All conditions have been met for assuming that t-procedures give close enough approximations (note that n = 200 is large). n = 200, _ y = 13.55, s = 3.1 1. H 0 : μ = 14 H A : < 14 (a left-tailed test) 3. Test Statistic: _ 13.55 14 2.053 / 3.1/ 200 y t s n - - = = = - 4. p-value = tcdf(-999, -2.053, 199) = 0.021 From Table T (with df = 180), the p-value is between 0.01 and 0.025 5. Reject H 0 at α = 0.10 and at = 0.05, but not at = 0.01 6. With a p-value of 0.021 and a 5% level of significance, the sample data provides enough evidence to conclude that, on average, adults now spend fewer than 14 hours per weekend on chores. 95% Confidence Interval (t* = 1.973) – formula on page 590: (13.12, 13.98) hours Based on this sample, we are 95% confident that, on average, adults now spend between 13.12 hours and 13.98 hours per weekend on chores. 2. Use 2-proportion z-procedures (chapter 22) because the sample data was collected from two independent populations (all men & all women) and the researcher wishes to compare the proportion of success between these populations. See the assumptions/conditions on pg. 560. All conditions have been met for assuming the distribution of ^ ^ 1 2 p p - is approximately normal. Men: ^ ^ 1 165 0.33 500 M p p = = = Women: ^ ^ 2 71 0.237 300 W p p = = = 1. H 0 : p M = p W or H 0 : p M - p W = 0 H A : p M > p W H A : p M - p W > 0 (a right-tailed test) 3. Test Statistic: ^ ^ ^ ^ ^ ^ 0.33 0.237 2.877 0.33 0.67 0.237 0.763 500 300 M W W W M M M W p p z p q p q n n - - = = = + + 4. From Z-Table: p(z 2.88) = 0.9980, so the p-value = 1 – 0.9980 = 0.0020 5. Reject H 0 at = 0.10, = 0.05, and = 0.01 6. With a p-value of 0.0020, the sample data provides very strong evidence to conclude that the percent of all men who regularly play the lottery is higher than the percent of all women who regularly play the lottery. 95% Confidence Interval (z* = 1.960) – formula on page 561: (0.030, 0.156), or 3.0% to 15.6% Based on this sample, we can say, with 95% confidence, the percent of all men who regularly play the lottery is between 3.0% and 15.6% higher than the percent of all women who regularly play the lottery.
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2 3. Use 1-sample t-procedures (chapter 23) because the data is a SRS from a single population (all packages of Sunlight Vanilla Wafers), the population has a normal distribution, and σ is unknown. Since the population’s distribution is known to be normal, t-procedures will give perfectly accurate
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InferentialStats-PracticeProblems-Answer - STA2023...

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