MATH 225N Week 7 Assignment (30 Q/A)/ MATH225 Week 7 Assignment: Hypothesis Test for the mean-Polution Standard Deviation known (Latest, 2020): Chamberlain College of Nursing | 100 % VERIFIED ANSWERS, GRADE A - $25.49   Add to cart

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MATH 225N Week 7 Assignment (30 Q/A)/ MATH225 Week 7 Assignment: Hypothesis Test for the mean-Polution Standard Deviation known (Latest, 2020): Chamberlain College of Nursing | 100 % VERIFIED ANSWERS, GRADE A

MATH225N Week 7 Assignment: Hypothesis Test for the mean-Polution Standard Deviation known MATH225 Week 7 Assignment: Hypothesis Test for the mean-Polution Standard Deviation known Compute the value of the test statistic (z-value) for a hypothesis test for one population mean with a known standard deviation Question Jamie, a bowler, claims that her bowling score is less than 168 points, on average. Several of her teammates do not believe her, so she decides to do a hypothesis test, at a 1% significance level, to persuade them. She bowls 17 games. The mean score of the sample games is 155 points. Jamie knows from experience that the standard deviation for her bowling score is 19 points. • H0: μ≥168; Ha: μ<168 • α=0.01 (significance level) What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places? ________________________________________ Provide your answer: Test statistic =-2.82 Correct answers: • Test statistic = −2.82 The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean score, x¯=155. The sample the bowler uses is 17 games, so n=17. She knows the standard deviation of the games, σ=19. Lastly, the bowler is comparing the population mean score to 168points. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=155−√≈−134.608≈−2.82 So, the test statistic for this hypothesis test is z0=−2.82. Distinguish between one- and two-tailed hypotheses tests and understand possible conclusions Question Which graph below corresponds to the following hypothesis test? H0:μ≥5.9, Ha:μ<5.9 Answer Explanation Correct answer: A normal curve is over a horizontal axis and is centered on 5.9. A vertical line segment extends from the horizontal axis to the curve at a point to the left of 5.9. The area under the curve to the left of the point is shaded. The alternative hypothesis, Ha, tells us which area of the graph we are interested in. Because the alternative hypothesis is μ<5.9, we are interested in the region less than (to the left of) 5.9, so the correct graph is the first answer choice. Your answer: wrong Identify the null and alternative hypotheses Question A politician claims that at least 68% of voters support a decrease in taxes. A group of researchers are trying to show that this is not the case. Identify the researchers' null hypothesis, H0, and the alternative hypothesis, Ha, in terms of the parameter p. ________________________________________ Select the correct answer below: ________________________________________ H0: p≤0.68; Ha: p>0.68 H0: p<0.68; Ha: p≥0.68 H0: p>0.68; Ha: p≤0.68 H0: p≥0.68; Ha: p<0.68 Perform and interpret a hypothesis test for a proportion using Technology - Excel Question Steve listens to his favorite streaming music service when he works out. He wonders whether the service's algorithm does a good job of finding random songs that he will like more often than not. To test this, he listens to 50 songs chosen by the service at random and finds that he likes 32 of them. Use Excel to test whether Steve will like a randomly selected song more often than not, and then draw a conclusion in the context of the problem. Use α=0.05. ________________________________________ Select the correct answer below: ________________________________________ Reject the null hypothesis. There is sufficient evidence to conclude that Steve will like a randomly selected song more often than not. Reject the null hypothesis. There is insufficient evidence to conclude that Steve will like a randomly selected song more often than not. Fail to reject the null hypothesis. There is sufficient evidence to conclude that Steve will like a randomly selected song more often than not. Fail to reject the null hypothesis. There is insufficient evidence to conclude that Steve will like a randomly selected song more often than not. Great work! That's correct.Correct answer: Reject the null hypothesis. There is sufficient evidence to conclude that Steve will like a randomly selected song more often than not. Step 1: The sample proportion is pˆ=3250=0.64, the hypothesized proportion is p0=0.5, and the sample size is n=50. Step 2: The test statistic, rounding to two decimal places, is z=0.64−0.50.5(1−0.5)50‾‾‾‾‾‾‾‾‾‾‾‾√≈1.98. Step 3: Since the test is right-tailed, entering the function =1−Norm.S.Dist(1.98,1) into Excel returns a p-value, rounding to three decimal places, of 0.024. Step 4: Since the p-value is less than α=0.05, reject the null hypothesis. There is sufficient evidence to conclude that Steve will like a randomly selected song more often than not. Null and alternative hypothesis: H0: p = 0.5 Ha: p >= 0.5 Level of Significance =0.05 Proportion under H0 = 0.5 n = 50 Number of Successes=32 Sample Proportion 0.64000 StDev 0.50000 SE 0. Test Statistic (z) 1. One-Sided p-value 0. Two-Sided p-value 0. P value = 0.02 < alpha (0.05) Hence, reject the null hypothesis. Perform and interpret a hypothesis test for a proportion using Technology - Excel Question A magazine regularly tested products and gave the reviews to its customers. In one of its reviews, it tested two types of batteries and claimed that the batteries from Company A outperformed the batteries from Company B. A representative from Company B asked for the exact data from the study. The author of the article told the representative from Company B that in 200 tests, a battery from Company A outperformed a battery from Company B in 108 of the tests. Company B decided to sue the magazine, claiming that the results were not significantly different from 50% and that the magazine was slandering its good name. Use Excel to test whether the true proportion of times that Company A's batteries outperformed Company B's batteries is different from 0.5. Identify the p-value, rounding to three decimal places. ________________________________________ Provide your answer below: ________________________________________ 0.258 p-value= Well done! You got it right.Correct answers: • p-value=0.258 Step 1: The sample proportion is pˆ==0.54, the hypothesized proportion is p0=0.5, and the sample size is n=200. Step 2: The test statistic, rounding to two decimal places, is z=0.54−0.50.5(1−0.5)200‾‾‾‾‾‾‾‾‾‾‾‾√≈1.13. Step 3: Since the test is two-tailed and the test statistic is positive, entering the function =2∗(1−Norm.S.Dist(1.13,1)) into Excel returns a p-value, rounding to three decimal places, of 0.258. Perform and interpret a hypothesis test for a proportion using Technology - Excel Question A candidate in an election lost by 5.8% of the vote. The candidate sued the state and said that more than 5.8% of the ballots were defective and not counted by the voting machine, so a full recount would need to be done. His opponent wanted to ask for the case to be dismissed, so she had a government official from the state randomly select 500 ballots and count how many were defective. The official found 45 defective ballots. Use Excel to test if the candidate's claim is true and that more than 5.8% of the ballots were defective. Identify the p-value, rounding to three decimal places. Well done! You got it right.Correct answers: • p-value=0.001 Step 1: The sample proportion is pˆ=45500=0.09, the hypothesized proportion is p0=0.058, and the sample size is n=500. Step 2: The test statistic, rounding to two decimal places, is z=0.09−0.0580.058(1−0.058)500‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√≈3.06. Step 3: Since the test is right-tailed, enter the function =Norm.S.Dist(3.06,0) into Excel and subtract from 1. This returns a p-value, rounding to three decimal places, of 0.001 Perform and interpret a hypothesis test for a proportion using Technology - Excel Question Dmitry suspected that his friend is using a weighted die for board games. To test his theory, he wants to see whether the proportion of odd numbers is different from 50%. He rolled the die 40 times and got an odd number 14 times. Dmitry conducts a one-proportion hypothesis test at the 5% significance level, to test whether the true proportion of odds is different from 50%. (a) Which answer choice shows the correct null and alternative hypotheses for this test? ________________________________________ Select the correct answer below: ________________________________________ H0:p=0.35; Ha:p>0.35, which is a right-tailed test. H0:p=0.5; Ha:p<0.5, which is a left-tailed test. H0:p=0.35; Ha:p≠0.35, which is a two-tailed test. H0:p=0.5; Ha:p≠0.5, which is a two-tailed test. Well done! You got it right. Correct answer: H0:p=0.5; Ha:p≠0.5, which is a two-tailed test. The null hypothesis should be true proportion: H0:p=0.5. Dmitry wants to know if the true proportion is different from 0.5. This means that we just want to test if the proportion is not 0.5. So, the alternative hypothesis is Ha:p≠0.5, which is a two-tailed test. PART 2 Perform and interpret a hypothesis test for a proportion using Technology - Excel Question Dmitry suspects that his friend is using a weighted die for board games. To test his theory, he wants to see whether the proportion of odd numbers is different from 50%. He rolled the die 40 times and got an odd number 14 times. Dmitry conducts a one-proportion hypothesis test at the 5% significance level, to test whether the true proportion of odds is different from 50%. (a) H0:p=0.5; Ha:p≠0.5, which is a two-tailed test. (b) Use Excel to test whether the true proportion of odds is different from 50%. Identify the test statistic, z, and p-value from the Excel output, rounding to three decimal places. test statistic=-1.897 p-value=0.058 PART 3 Perform and interpret a hypothesis test for a proportion using Technology - Excel Question Dmitry suspects that his friend is using a weighted die for board games. To test his theory, he wants to see whether the proportion of odd numbers is different from 50%. He rolled the die 40 times and got an odd number 14 times. Dmitry conducts a one-proportion hypothesis test at the 5% significance level, to test whether the true proportion of odds is different from 50%. (a) H0:p=0.5; Ha:p≠0.5, which is a two-tailed test. (b) z0=−1.897, p-value is = 0.058 (c) Which of the following are appropriate conclusions for this hypothesis test? Select all that apply. ________________________________________ Select all that apply: ________________________________________ • We reject H0. • We fail to reject H0. • At the 5% significance level, the data provide sufficient evidence to conclude the true proportion of odds is different than 50%. • At the 5% significance level, the data do not provide sufficient evidence to conclude the true proportion of odds is different than 50%. • option 2 and 4 are answer • Explanation: • (c) From part b, p-value is = 0.058 > 0.05 significance level • So, we failed to reject the null hypothesis Ho • • And • At the 5% significance level, the data do not provide sufficient evidence to conclude the true proportion of odds is different than 50%. Identify the null and alternative hypotheses Question A politician claims that at least 68% of voters support a decrease in taxes. A group of researchers are trying to show that this is not the case. Identify the researchers' null hypothesis, H0, and the alternative hypothesis, Ha, in terms of the parameter p. ________________________________________ Select the correct answer below: ________________________________________ H0: p≤0.68; Ha: p>0.68 H0: p<0.68; Ha: p≥0.68 H0: p>0.68; Ha: p≤0.68 H0: p≥0.68; Ha: p<0.68 4-H0: p≥0.68; Ha: p<0.68 Explanation: First, the equal condition is written in the null hypothesis. Second, the statement that at least 68% of taxes support a decrease in taxes is expressed as: H0:P≥0.68 second, the alternative hypothesis is the investigation hypothesis, the opposite of the null hypothesis: Ha:P<0.68 Third, therefore, the hypotheses are H0:P≥0.68 ; Ha:P<0.68 Well done! You got it right.Let the parameter p be used to represent the proportion. The null hypothesis is always stated with some form of equality: equal (=), greater than or equal to (≥), or less than or equal to (≤). Therefore, in this case, the null hypothesis H0 is p≥0.68. The alternative hypothesis is contradictory to the null hypothesis, so Hais p<0.68. Also, remember that the alternative hypothesis is the statement that the research or study is trying to show. In this case, they are trying to show that the politician is wrong, so the alternative hypothesis is the opposite of what the politician said. So Ha is p<0.68. The null hypothesis is what the politician said, so H0 is p≥0.68. Differentiate between Type I and Type II errors when performing a hypothesis test Question Determine the Type II error if the null hypothesis, H0, is: a wooden ladder can withstand weights of 250pounds and less. ________________________________________ Select the correct answer below: ________________________________________ You think the ladder can withstand weight of 250 pounds and less when, in fact, it cannot. You think the ladder cannot withstand weight of 250 pounds and less when, in fact, it really can. You think the ladder can withstand weight of 250 pounds and less when, in fact, it can. You think the ladder cannot withstand weight of 250 pounds and less when, in fact, it cannot. Perfect. Your hard work is paying off

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