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

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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. Great work! That's correct.

Answer Explanation Correct answers:

• (^) 0.08 4 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. 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.
• (^) p=0. 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

Marty, a typist, claims that his average typing speed is 72 words per minute. During a practice session, Marty has a sample typing speed mean of 84 words per minute based on 12 trials. At the 5% significance level, does the data provide sufficient evidence to conclude that his mean typing speed is greater than 72 words per minute? Accept or reject the hypothesis given the sample data below.

• H0:μ≤72 words per minute; Ha:μ>72 words per minute
• α=0.05 (significance level)
• z0=2.
• p=0.018 Select the correct answer below: Do not reject the null hypothesis because 2.1>0.05. Do not reject the null hypothesis because the value of z is positive. Reject the null hypothesis because 2.75>0.05. Reject the null hypothesis because the p-value 0.018 is less than the significance level α=0.05. Do not reject the null hypothesis because the p-value 0.018 is less than the significance level α=0.05. Perfect. Your hard work is paying off 😀 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=0.27? (Do not round your answer; compute your answer using a value from the table below.) z0.10.20.30.40.5............ Well done! You got it right.

The p-value is the probability of an observed value of z=0.27 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=−0.27, or greater than z=0.27. This probability is equal to the area under the Standard Normal curve that lies either to the left of z=−0.27, or to the right of z=0.27.

A standard normal curve with three points labeled on the horizontal axis labeled z. The mean is labeled at 0 and observed values of negative 0.27 and 0.27 are labeled. The areas under the curve and to the left of negative 0.27 and to the right of 0.27 are shaded. The shaded areas are both labeled p-value. Using the Standard Normal Table, we can see that the p-value that corresponds with z=0.27 is 0.606, which is the area to the left of z=0.27. However, we want the area to the right of 0.27, which is 1−0.606=0.394. Because the Standard Normal curve is symmetric, the area to the left of z=−0.27 is 0.394 as well. So, the p-value of this two-tailed one-mean hypothesis test is (2)(0.394)=0.788. Determine the p-value for a hypothesis test for the mean (population standard deviation known) Question Kurtis is a statistician who claims that the average salary of an employee in the city of Yarmouth is no more than $55,000 per year. Gina, his colleague, believes this to be incorrect, so she randomly selects 61employees who work in Yarmouth and record their annual salary. Gina calculates the sample mean income to be $56,500 per year with a sample standard deviation of 3,750. Using the alternative hypothesis Ha:μ>55,000, find the test statistic t and the p-value for the appropriate hypothesis test. Round the test statistic to two decimal places and the p-value to three decimal places. Right-Tailed T-Table probability 0.0004 0.0014 0.0024 0.0034 0.0044 0.0054 0. Degrees of Freedom 54 3.562 3.135 2.943 2.816 2.719 2.641 2. 55 3.558 3.132 2.941 2.814 2.717 2.640 2.

Yes that's right. Keep it up! t =3.12, p-value=0. Answer Explanation Correct answers:

• (^) t =3.12, p-value=0.00 1 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. 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. 57 3.550 3.127 2.937 2.810 2.714 2.636 2. 58 3.547 3.125 2.935 2.808 2.712 2.635 2. 59 3.544 3.122 2.933 2.806 2.711 2.633 2. 60 3.540 3.120 2.931 2.805 2.709 2.632 2. probability 0.0004 0.0014 0.0024 0.0034 0.0044 0.0054 0.

Correct! You nailed it.

Answer Explanation Correct answers:

• (^) 0.11 2 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.
• p=0. 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.
• p=0. 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.

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=0.0721, 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 Ruby, a bowler, has a sample game score mean of 125.8 from 25 games. Ruby still claims that her average game score is 140, and the low average can be attributed to chance. At the 5% significance level, does the data provide sufficient evidence to conclude that Ruby's mean game score is less than 140? Given the sample data below, accept or reject the hypothesis.

• (^) H0:μ=140; Ha:μ<
• (^) α=0.05 (significance level)
• (^) z0=−0.
• (^) p=0. Select the correct answer below: Reject the null hypothesis because the value of z is negative. Do not reject the null hypothesis because |−0.52|>0.05. Reject the null hypothesis because the p-value 0.3015 is greater than the significance level α=0.05. Do not reject the null hypothesis because the p-value 0.3015 is greater than the significance level α=0.05. Reject the null hypothesis because |−0.52|>0.05. Do not reject the null hypothesis because the p-value 0.3015 is greater than the significance level α=0. Explanation: Decision rule using the P −value and significance level.If P −value< α : Reject H 0 .If P −value> α : Fail to reject H 0.

Decision using the P −value and significance level.We compare the P −value with the s ignificance level ( α ). P −value (0.3015000000)> α (0.05)Decision: Fail to reject H 0. Great work! That's correct. Correct answer: Do not reject the null hypothesis because the p-value 0.3015 is greater than the significance level α=0.05. 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=0.05 is less than or equal to p=0.3015, so the decision is to not reject the null hypothesis. Marie, a bowler, has a sample game score mean of 129.2 from 24 games. Marie still claims that her average game score is 143, and the low average can be attributed to chance. At the 5% significance level, does the data provide sufficient evidence to conclude that Marie's mean game score is less than 143? Given the sample data below, accept or reject the hypothesis.

• (^) H0:μ=143; Ha:μ<
• (^) α=0.05 (significance level)
• (^) z0=−1.
• (^) p=0.
• (^) Accept the null hypothesis
• (^) No,at 0.05 significance level, the data does not provide sufficient evidence to conclude that Marie's mean game score is less than 143.
• Explanation:
• (^) Ho : μ =143 Ha : μ <
• (^) values are given as ,
• (^) α=0.05 (significance level)
• (^) z-statistic =−1.
• (^) p-value=0.
• (^) [0.1423 > 0.05] p value is greater than alpha ,so accept the null hypothesis
• (^) No,at 0.05 significance level, the data does not provide sufficient evidence to conclude that Marie's mean game score is less than 143.

Perfect. Your hard work is paying off 😀. Correct answer: Do not reject the null hypothesis because the p-value 0.1423 is greater than the significance level α=0.05. 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=0.05 is less than or equal to p=0.1423, so the decision is to not reject the null hypothesis.