MATH 225N Week 7 Assignment Conducting a Hypothesis Test for Mean – Population Standard De, Exams of Mathematics

Download MATH 225N Week 7 Assignment Conducting a Hypothesis Test for Mean – Population Standard De and more Exams Mathematics in PDF only on Docsity! MATH 225N Week 7 Assignment Conducting a Hypothesis Test for Mean – Population Standard Deviation Known P-Value Approach Question What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=−1.73? (Do not round your answer; compute your answer using a value from the table below.) z−1.8−1.7−1.6−1.5−1.4 Great work! That's correct. 0.084 Answer Explanation Correct answers: • 0.084 The p-value is the probability of an observed value of z=1.73 or greater in magnitude if the null hypothesis is true, because this hypothesis test is two-tailed. This means that the p-value could be less than z=−1.73, or greater than z=1.73. This probability is equal to the area under the Standard Normal curve that lies either to the left of z=−1.73, or to the right of z=1.73. A normal curve is over a horizontal axis and is centered on 0. Two points are labeled negative 1.73 and 1.72 The area to the right of 1.73 and to the left of negative 1.73 is shaded. Using the Standard Normal Table given, we can see that the p-value that corresponds with z=−1.73 is 0.042, which is just the area to the left of z=−1.73. Since the Standard Normal curve is symmetric, the area to the right of z=1.73 is 0.042 as well. So, the p- value of this two-tailed one-mean hypothesis test is (2)(0.042)=0.084 Make a conclusion and interpret the results of a one-mean hypothesis test (population standard deviation known) using the P-Value Approach Question Mary, a javelin thrower, claims that her average throw is 61 meters. During a practice session, Mary has a sample throw mean of 55.5 meters based on 12 throws. At the 1% significance level, does the data provide sufficient evidence to conclude that Mary's mean throw is less than 61 meters? Accept or reject the hypothesis given the sample data below. • H0:µ=61 meters; Ha:µ<61 meters • α=0.01 (significance level) • z0=−1.99 • p=0.0233 Great work! That's correct. Reject the null hypothesis because |−1.99|>0.01. Do not reject the null hypothesis because |−1.99|>0.01. Reject the null hypothesis because the p-value 0.0233 is greater than the significance level α=0.01. Do not reject the null hypothesis because the value of z is negative. Do not reject the null hypothesis because the p-value 0.0233 is greater than the significance level α=0.01. Answer Explanation Correct answer: Do not reject the null hypothesis because the p-value 0.0233 is greater than the significance level α=0.01. In making the decision to reject or not reject H0, if α>p-value, reject H0 because the results of the sample data are significant. There is sufficient evidence to conclude that H0 is an incorrect belief and that the alternative hypothesis, Ha, may be correct. If α≤p-value, do not reject H0. The results of the sample data are not significant, so there is not sufficient evidence to conclude that the alternative hypothesis, Ha, may be correct. In this case, α=0.01 is less than or equal to p=0.0233, so the decision is to not reject the null hypothesis. Make a conclusion and interpret the results of a one-mean hypothesis test (population standard deviation known) using the P-Value Approach Question Yes that's right. Keep it up! t=3.12, p-value=0.001 Answer Explanation Correct answers: • t=3.12, p-value=0.001 Since σ is unknown and the sample size is at least 30, the hypothesis test for the mean can be performed using the t-distribution. Here, the sample mean x⎯⎯⎯ is 56,500, the hypothesized mean µ is 55,000, the sample standard deviation s is 3,750, and the sample size n is 61. Substitute these values into the formula to calculate the t test statistic. tt=56,500−55,0003,750/61‾‾‾√≈3.12 Now find the p-value. Notice that the test statistic has 61−1=60 degrees of freedom and that this is a right-tailed test because the alternative hypothesis is Ha:µ>55,000. Find the p-value for a right-tailed test of a t-distribution with 60 degrees of freedom, where t≈3.12. That is, to find the p-value, find the area under the t-distribution curve with 60 degrees of freedom to the right of t≈3.12. The p-value that corresponds to these conditions is approximately 0.001. Determine the p-value for a hypothesis test for the mean (population standard deviation known) Question What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=−1.59? (Do not round your answer; compute your answer using a value from the table below.) z−1.8−1.7−1.6−1.5−1.40...... 56 3.554 3.130 2.939 2.812 2.716 2.638 2.572 57 3.550 3.127 2.937 2.810 2.714 2.636 2.571 58 3.547 3.125 2.935 2.808 2.712 2.635 2.569 59 3.544 3.122 2.933 2.806 2.711 2.633 2.568 60 3.540 3.120 2.931 2.805 2.709 2.632 2.567 probability 0.0004 0.0014 0.0024 0.0034 0.0044 0.0054 0.0064 Correct! You nailed it. 0.112 Answer Explanation Correct answers: • 0.112 The p-value is the probability of an observed value of z=1.59 or greater in magnitude if the null hypothesis is true, because this hypothesis test is two-tailed. This means that the p-value could be less than z=−1.59, or greater than z=1.59. This probability is equal to the area under the Standard Normal curve that lies either to the left of z=−1.59, or to the right of z=1.59. A normal curve is over a horizontal axis and is centered on 0.00 with ticks at negative 1.59 and 1.59. The area under the curve and to the left of negative 1.59 and to the right of 1.59 is shaded. Using the Standard Normal Table given, we can see that the p-value that corresponds with z=−1.59 is 0.056, which is just the area to the left of z=−1.59. Since the Standard Normal curve is symmetric, the area to the right of z=1.59 is 0.056 as well. So, the p- value of this two-tailed one-mean hypothesis test is (2)(0.056)=0.112. Make a conclusion and interpret the results of a one-mean hypothesis test (population standard deviation known) using the P-Value Approach Question Nancy, a golfer, claims that her average driving distance is 253 yards. During a practice session, Nancy has a sample driving distance mean of 229.6 yards based on 18 drives. At the 2% significance level, does the data provide sufficient evidence to conclude that Nancy's mean driving distance is less than 253 yards? Accept or reject the hypothesis given the sample data below. •H0:µ=253 yards; Ha:µ<253 yards •α=0.02 (significance level) •z0=−0.75 •p=0.2266 Perfect. Your hard work is paying off 😀 Correct answer: Do not reject the null hypothesis because the p-value 0.2266 is greater than the significance level α=0.02. In making the decision to reject or not reject H0, if α>p-value, reject H0 because the results of the sample data are significant. There is sufficient evidence to conclude that H0 is an incorrect belief and that the alternative hypothesis, Ha, may be correct. If α≤p-value, do not reject H0. The results of the sample data are not significant, so there is not sufficient evidence to conclude that the alternative hypothesis, Ha, may be correct. In this case, α=0.02 is less than or equal to p=0.2266, so the decision is to not reject the null hypothesis. Make a conclusion and interpret the results of a one-mean hypothesis test (population standard deviation known) using the P-Value Approach Question Kathryn, a golfer, has a sample driving distance mean of 187.3 yards from 13 drives. Kathryn still claims that her average driving distance is 207 yards, and the low average can be attributed to chance. At the 1%significance level, does the data provide sufficient evidence to conclude that Kathryn's mean driving distance is less than 207 yards? Given the sample data below, accept or reject the hypothesis. •H0:µ=207 yards; Ha:µ<207 yards •α=0.01 (significance level) •z0=−1.46 •p=0.0721 Keep trying - mistakes can help us grow. Correct answer: Do not reject the null hypothesis because the p-value 0.0721 is greater than the significance level α=0.01.