MATH 225N Week 7 Hypothesis Testing Questions and Answers (MATH225N)

Exam (elaborations) MATH 225N Week 7 Hypothesis Testing Questions and Answers (MATH225N) Week 7 Hypothesis Testing Q & A 1. Steve listens to his favorite streaming music service when he works out. He wonders whether the service 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 than not and then draw a conclusion in the context of a problem. Use α = 0.05. Type equation here . Ho: p = ≤ 0.5 (50%) p = 0.5 Ha: p = > 0.5 (strictly ¿≠ ) P-value = 0.02 which is < α=0.05 we reject Ho and support the Ha Hypothesis Test for p population proportion Level of Significance 0.05 (decimal ) Proportion under H0 0.5000 (decimal ) n 50 Number of Successes 32 Sample Proportion 0.64000 0 StDev 0.50000 0 SE 0.07071 1 Test Statistic (z) 1.97989 9 One-Sided p-value 0.02385 2 Two-Sided p-value 0.04770 4 Right-Tailed (>) 1. Left-Tailed (<) -1. Two-Tailed (≠) ± 1. Answer: Reject the null hypothesis. There is sufficient evidence to prove that Steve will like a random selected song more often than not. 2. A magazine regularly tested products and gave the reviews to its customers. In one of its reviews, it tested 2 types of batteries and claimed that the batteries from company A outperformed batteries from company B in 108 of the tests. There were 200 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 it to 3 decimal places. Ho: p = 0.5 Ha ≠ 0.5 (two tailed test) n = 200 (α is not given soleave it 0.05) Hypothesis Test for p population proportion Level of Significance 0.05 Proportion under H0 0.5000 n 200 Number of Successes 108 Sample Proportion 0. StDev 0. SE 0. Test Statistic (z) 1. One-Sided p-value 0. Two-Sided p-value 0. Right-Tailed (>) 1. Left-Tailed (<) - 1. Two-Tailed (≠) ± 1. Answer: 0.258 (because it is a two tailed test). We are not rejecting the null hypothesis and we do not have evidence to support the alternative hypothesis. 3. 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 21 defective ballots. Use Excel to test if the candidates claim is true and that < 5.8% of the ballots were defective. Identify the p=value rounding to 3 decimal places. Ho: p = ≥ 0.058 Ha ¿0.058 (one tailed test) n = 500 (α is not given soleave it 0.05) Hypothesis Test for p population proportion Level of Significance 0.05 (decimal) Proportion under H0 0.0580 (decimal) n 500 Number of Successes 21 Sample Proportion 0. StDev 0. SE 0. Test Statistic (z) - 1. One-Sided p-value 0. Two-Sided p-value 0. Right-Tailed (>) 1. Left-Tailed (<) -1. Two-Tailed (≠) ± 1. Answer: 0.063 4. A researcher claims that the incidence of a certain type of cancer is < 5%. To test this claim, a random sample of 4000 people are checked and 170 are found to have the cancer. The following is the set up for the hypothesis: Ho = 0.05 Ha = < 0.05 In the example the p-value was determined to be 0.015. Come to a conclusion and interpret the results of this hypothesis test for a proportion (use a significance level of 5%) Answer: The decision is to reject the null hypothesis. The conclusion is that there is enough evidence to support the claim. 5. A researcher is investigating a government claim that the unemployment rate is < 5%. TO test this claim, a random sample of 1500 people is taken and it is determined that 61 people were unemployed. Ho: p = 0.05 Ha: p < 0.05 Find the p-value for this hypothesis test for a proportion & round to 3 decimal places. Hypothesis Test for p population proportion Level of Significance 0.05 Proportion under H0 0.0500 n 1500 Number of Successes 61 Sample Proportion 0.04066 7 StDev 0.21794 5 SE 0.00562 7 Test Statistic (z) - 1.65857 7 One-Sided p-value 0.04845 7 Two-Sided p-value 0.09691 4 Answer: 0.048 6. An economist claims that the proportion of people that plan to purchase a fully electric vehicle as their next car is greater than 65%. To test this claim, a random sample of 750 people were asked if they planned to purchase a fully electric vehicle as their next car. Of this 750, 513 indicated that they plan to purchase an electric vehicle. Ho: p = 0.65 Ha; p = >0.65 Find the p-value for this hypothesis test for a proportion & round to 3 decimal places. Hypothesis Test for p population proportion Level of Significance 0.05 Proportion under H0 0.6500 n 750 Number of Successes 513 Sample Proportion 0.68400 0 StDev 0.47697 0 SE 0.01741 6 Test Statistic (z) 1.95217 5 One-Sided p-value 0.02558 8 Two-Sided p-value 0.05117 6 Answer: 0.026 7. Colton makes the claim to his classmates that < 50% of newborn babies born this year in his state are boys. To prove this claim, he selects a random sample of 344 birth records in his state from this year. Colton found that 176 of the newborns were boys. What are the null and alternative hypothesis for this hypothesis test. Answer: Ho: 0.5 Ha: <0.5 8. An Airline company claims that in its recent advertisement that at least 94% of passenger luggage that is lost is recovered and reunited with their customer within 1 day. Hunter is a graduate student studying statistics. For a research project, Hunter wants to find out whether there is sufficient evidence in support of the airline company’s claim. He randomly selects 315 passengers whose luggage was lost by the airlines and found out that 276 of those passengers were reunited with their luggage within 1 day. Are all of the conditions for his hypotheses test met, and if so, what are the Ho and Ha for this hypothesis test? For a binomial Model to follow the normal model, the following condition must be satisfied: Success count = n * p ≥ 5 and Failure count ≥5 Example: success count 315 * 0.94 = 296.1 and failure count 315-296.1 = 18.9 so it meets the conditions. Answer: All of the conditions were met and the Ho = 0.94; Ha = >0.94 9. A college administrator claims that the proportion of students who are nursing majors is > 40%. To test this claim, a group of 400 students are randomly selected and its determined that 190 are nursing majors. The following is the set up for the hypothesis test: Ho: p = .40 and Ha: p = >.40 Find the test statistics for this hypothesis test for a proportion & round to 2 decimal places. Answer: 3.06 Level of Significance 0.05 Proportion under H0 0.4000 n 400 Number of Successes 190 Sample Proportion 0. StDev 0. SE 0. Test Statistic (z) 3. One-Sided p-value 0. Two-Sided p-value 0. 10. A hospital administrator claims that the proportion of knee surgeries that are successful are 87%. To test this claim, a random sample of 450 patients who underwent knee surgery is taken and it is determined that 371 patients had a successful knee surgery operation. Ho: p = 0.87 Ha: p ≠ 0.87 (two sided tail) Find the test statistics for this hypothesis test for a proportion & round to 2 decimal places. Answer: -2.87 (this would be rejected) Level of 0.05 Significance Proportion under H0 0.8700 n 450 Number of Successes 371 Sample Proportion 0. StDev 0. SE 0. Test Statistic (z) -2. One-Sided p-value 0. Two-Sided p-value 0. 11. Jose, a competitor in cup stacking, has a sample stacking time mean of 7.5 seconds from 13 trials. Jose still claims that his average stacking time is 8.5 seconds, and the low average can be contributed to chance. At the 2% significant level, does the data provide sufficient evidence to conclude that Jose’s mean stacking time is less than 8.5 seconds? Given the sample data below, select or reject the hypothesis. (If p=value is < alpha value, we would automatically reject the hypothesis) Ho: μ = 8.5 Ha: μ = <8.5 α = 0.02 (significance level) Zo = -2.18 P = 0.0146 Answer: Reject the null hypothesis because the p value 0.0146 is less than the significance level 0.02 12. Marty, a typist, claims his average typing speed is 72 wpm. During a practice session, Marty has a sample typing speed mean of 84 wpm based on 12 trials. At the 5% significance level, does the data provide sufficient evidence to conclude that his mean typing speed is >72 wpm? Accept or reject the hypothesis given the data below. Ho: μ=72wpm ; Ha: μ=¿72wpm; α=0.05 (significance level) ; Zo = 2.1; p = 0.018 Answer: Reject the null hypothesis because the p-value 0.018 is less than the