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MATH 225N Week 7 Hypothesis Testing Question and Answers 2024
- 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 often. 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. Ha: p = > 0.5 (strictly ¿> ≠^ ) P-value = 0.02 which is < Hypothesis Test for p population proportion α =0.05 we reject Ho and support the Ha Level of (decimal) Significance Proportion under
(decimal) H (^0) 0. n Number of
Successes 32 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: Reject the null hypothesis. There is sufficient evidence to prove that Steve will like a random selected song more often than not.
- 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. Proportion under H 0 0. n 200 Number of Successes 108 Two-Sided p-value Right-Tailed (>) Left-Tailed (<) Two-Tailed (≠) ± Sample Proportion StDev 0. SE 0. Test Statistic (z) 0. One-Sided p-value 1. 0 0.129238. 25847 6 1.644854-
- 64485 4
- 95996 4 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.
- 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,
St Dev SE Test Statistic (z) One-Sided p-value Two-Sided p-value Sample Proportion 0. 0
- 2337 4 3
- 0104 5 3
Right Tailed (>) 1.
Left-Tailed (<) 1. Two-Tailed (≠) ± 1. Answer: 0.
- 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. Ha = <
- 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.
- 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. Find the p-value for this hypothesis test for a proportion & round to 3 decimal places.
Hypothesis Test for p population proportion
- 0500 1500 Level of Significance Proportion under H 0 n Number of Successes One-Sided p- value Two-Sided p- value
Sample Proportion 0. 7 StDev 0. 5 SE 0. 7
Test Statistic (z) 1.^658757
7
4 Answer: 0.
- (^) 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. Find the p-value for this hypothesis test for a proportion & round to 3 decimal places. Hypothesis Test for p population proportion
- 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 = >. Find the test statistics for this hypothesis test for a proportion & round to 2 decimal places. Answer: 3. Level of Significance Proportion under H 0 n Number of Successes Sample Proportion
StDev 0. SE 0. Test Statistic (z) 3.^061862 One-Sided p- (^) 0. value Two-Sided p-
value
- (^) 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 Significance Proportion under H 0 n Nu mber of Successes
Sample Proportion StDev SE
Test Statistic (z) 2.^873534 One-Sided p- (^) 0. value Two-Sided p- (^) 0. value
- 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
- 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 > wpm? Accept or reject the hypothesis given the data below. Ho: μ = 72 wpm ; Ha: μ =¿ 72 wpm; α =0.05 ( significance level ) ; Zo = 2.1; p = 0. Answer: Reject the null hypothesis because the p-value 0.018 is less than the significance level α =0.
- What is the p-value of a right-tailed one mean hypothesis test, with a test statistic of Zo = 2.1? (Do not round your answer. Compute your answer using a value from the table. (Value in table was 0.982) 1 – 0.982 = p=value of 0. Answer: 0.
- (^) What is the p-value of a two-tailed one mean hypothesis test, with a test statistic of Zo = 0.27? (Do not round your answer. Compute your answer using a value from the table. (Value in table was 0.606) 1-0.606 = P-value of 0.394 because it is two tailed, you multiply 2 x .394 =. Answer: 0.
- (^) Raymond, a typist, claims his average typing speed is 89 wmp. During a practice session, Raymond has a sample typing speed mean of 95.5 wmp based on 15 trials. At the 1% significance level, does the data provide sufficient evidence to conclude that his mean typing speed is > 89 wmp? Accept or reject the hypothesis given the sample data below: Ho: μ = 89 ; Ha: μ =¿ 89 wmp α =0.02 Sig Level; Zo = 2.75; p = 0. Answer: Reject the null hypothesis because p-value 0.003 is less than alpha 0.
- (^) 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 61 employees 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 $3750. Using the alternative hypothesis, Ha = μ =¿ $ 55,000 , find the test statistic τ and the p-value for the appropriate hypothesis test. Round the τ to 2 decimal places and the p-value to 3 decimal places. Hypothesis Test for μ for unknown population stdev
M
Level of (decimal) Significance Mean 0. under H (^0 ) n Sample 61 Mean 56500 StDev 3750
MATH 225N Week
value Two-Sided p- value Answer: 0.
- A researcher claims that the incidence of a certain type of cancer is less than 5%. To test this claim, the a random sample of 4000 people are checked and 170 are determined to have the cancer. The following is the setup for this hypothesis test: H0:p=0. Ha:p<0. In this example, the p-value was determined to be 0.015. Come to a conclusion and interpret the results for this hypothesis test for a proportion (use a significance level of 5%) Select the correct answer below: The decision is to reject the Null Hypothesis. The conclusion is that there is enough evidence to support the claim. (p=0.015 α^ =0.05^ ¿ The decision is to fail to reject the Null Hypothesis. The conclusion is that there is not enough evidence to support the claim. (So, if p≤α, reject H0; otherwise fail to reject H0)
- (^) A police office claims that the proportion of people wearing seat belts is less than 65%. To test this claim, a random sample of 200 drivers is taken and its determined that 126 people are wearing seat belts. The following is the setup for this hypothesis test: H0:p=0. Ha:p<0. In this example, the p-value was determined to be 0.277.
The decision is to fail to reject the Null Hypothesis. The conclusion is that there is not enough evidence to support the claim.
- A police officer claims that the proportion of accidents that occur in the daytime (versus nighttime) at a certain intersection is 35%. To test this claim, a random sample of 500 accidents at this intersection was examined from police records it is determined that 156 accidents occurred in the daytime. The following is the setup for this hypothesis test: H 0: p = 0. Ha : p ≠ 0. Find the p-value for this hypothesis test for a proportion and round your answer to 3 decimal places. Provide your answer below: 0.
- A teacher claims that the proportion of students expected to pass an exam is greater than 80%. To test this claim, the teacher administers the test to 200 random students and determines that 151 students pass the exam. The following is the setup for this hypothesis test: H0:p= 0. 80 Ha:p> 0. 80 In this example, the p- value was determine d to be 0.944. Come to a conclusion and interpret the results for this hypothesis test for a proportion (use a significance level of 5%) The decision is to reject the Null Hypothesis. The conclusion is that there is enough evidence to support the claim. The decision is to fail to reject the Null Hypothesis. The conclusion is that there is not enough evidence to support the claim.
MATH 225N Week
MATH 225N
- (^) A researcher claims that the proportion of smokers in a certain city is less than 20%. To test this claim, a random sample of 700 people is taken in the city and 150 people indicate they are smokers. The following is the setup for this hypothesis test: H0:p=0.
MATH 225N Week 7 Hypothesis Testing Question and Answers
Latest Update 2023 Rated A+
MATH 225N Week 7 Hypothesis Testing Question and Answers
Latest Update 2023 Rated A+
Sample Proportion StDev SE Test Statistic (z) One-Sided p-
MATH 225N Week 7 Hypothesis Testing Question and Answers
Latest Update 2023 Rated A+
MATH 225N Week 7 Hypothesis Testing Question and Answers
Latest Update 2023 Rated A+
value Two-Sided p- 0. value
- 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. Select the correct answer below: Reject the null hypothesis because the p-value 0.0721 is greater than the significance level α=0.01. Do not reject the null hypothesis because the p-value 0.0721 is greater than the significance level α=0.01. Reject the null hypothesis because |−1.46|>0.01. Do not reject the null hypothesis because |−1.46|>0.01. Do not reject the null hypothesis because the value of z is negative.
- 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. Select the correct answer below: Reject the null hypothesis because |−1.99|>0.01.
MATH 225N Week 7 Hypothesis Testing Question and Answers
Latest Update 2023 Rated A+
