Using a 5% level of significance and a sample size of 24, what is the critical t-value for a null hypothesis, H0: µ ≤ 100?
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Q: Test the claim that the proportion of people who own cats is significantly different than 50% at the…
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Using a 5% level of significance and a
Given: Using a 5% level of significance and a sample size of 24, what is the critical t-value for a null hypothesis, H0: µ ≤ 100?
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- Kenneth, a competitor in cup stacking, claims that his average stacking time is 8.2 seconds. During a practice session, Kenneth has a sample stacking time mean of 7.8 seconds based on 11 trials. At the 4% significance level, does the data provide sufficient evidence to conclude that Kenneth's mean stacking time is less than 8.2 seconds? Accept or reject the hypothesis given the sample data below. H0:μ=8.2 seconds; Ha:μ<8.2 seconds α=0.04 (significance level) z0=−1.75 p=0.0401 Select the correct answer below: a. Do not reject the null hypothesis because the p-value 0.0401 is greater than the significance level α=0.04. b. Reject the null hypothesis because the p-value 0.0401 is greater than the significance level α=0.04. c. Reject the null hypothesis because the value of z is negative. d. Reject the null hypothesis because |−1.75|>0.04. e. Do not reject the null hypothesis because |−1.75|>0.04.Test the claim that the proportion of people who own cats is significantly different than 80% at the 0.1 significance level.The null and alternative hypothesis would be: H0:μ=0.8H0:μ=0.8H1:μ≠0.8H1:μ≠0.8 H0:p=0.8H0:p=0.8H1:p≠0.8H1:p≠0.8 H0:p≤0.8H0:p≤0.8H1:p>0.8H1:p>0.8 H0:p≥0.8H0:p≥0.8H1:p<0.8H1:p<0.8 H0:μ≤0.8H0:μ≤0.8H1:μ>0.8H1:μ>0.8 H0:μ≥0.8H0:μ≥0.8H1:μ<0.8H1:μ<0.8 The test is: right-tailed left-tailed two-tailed Based on a sample of 200 people, 87% owned catsThe p-value is: (to 2 decimals)Based on this we: Reject the null hypothesis Fail to reject the null hypothesisIn the past, the mean lifetime for a certain type of flashlight battery has been 9.5 hours. The manufacturer has introduced a change in the production method and wants to perform a significance test to determine whether the mean lifetime has increased as a result. The hypotheses are: null: μ = 9.5 hours alternative: u 9.5 hours Suppose that the results of a significance test result in a rejection of the null hypothesis. If we later find out that the true population running time was actually equal to 10.5, what error has been committed? Type I and Type II error Type II error no error Type I error
- A physical therapist is interested in studying the proportion of adults that receive a physical therapy treatment plan due to sciatica. The physical therapist states the proportion of adults that receive a physical therapy treatment plan due to sciatica is less than 0.30. If we would like to test the physical therapist's claim with a hypothesis test using a significance level of α=0.10 , which of the following choices are true? Select the correct answer below: There is a 10% chance we will conclude p=0.30, but is in fact p<0.30. There is a 10% chance of rejecting the null hypothesis. There is a 10% chance we will conclude p<0.30, but is in fact p=0.30. There is a 10% chance that p<0.30B. Identify the test statistic. ______ C. Identify the P-value. ____ D. What is the conclusion for this test? The P-value is (greater than/less than) the significance level α, so (reject/ fail to reject) the null hypothesis. There is (sufficient/ insufficient) evidence to support the claim that vinyl gloves have a greater virus leak rate than latex gloves.A researcher wants to test whether the mean height for men and women are different, using a significance level of 0.05. Which is the correct hypothesis? A Ho: P1 = P2 Ha: P1 P2 D Ho: P1 = μ2 Ha: H1 H2 O C A OF OE O B B Ho: P1 = P2 Ha: P1 P2 E Ho: P1 = μ2 Ha: 12 C Ho: P1 = P2 Ha: 1: P₁ P2 F H₂:₁ = 1₂ Ha: ₁ 2
- Test the claim that the proportion of people who own cats is smaller than 60% at the 0.005 significance level.The null and alternative hypothesis would be: H0:μ=0.6H0:μ=0.6H1:μ≠0.6H1:μ≠0.6 H0:p=0.6H0:p=0.6H1:p≠0.6H1:p≠0.6 H0:p≤0.6H0:p≤0.6H1:p>0.6H1:p>0.6 H0:μ≥0.6H0:μ≥0.6H1:μ<0.6H1:μ<0.6 H0:p≥0.6H0:p≥0.6H1:p<0.6H1:p<0.6 H0:μ≤0.6H0:μ≤0.6H1:μ>0.6H1:μ>0.6 The test is: two-tailed right-tailed left-tailed Based on a sample of 200 people, 51% owned catsThe test statistic is: (to 2 decimals)The p-value is: (to 2 decimals)Based on this we: Reject the null hypothesis Fail to reject the null hypothesisTest the claim that the proportion of people who own cats is significantly different than 80% at the 0.2 significance level. The null and alternative hypothesis would be: Ho: p= 0.8 Ho: ≤ 0.8 Hop: 0.8 Ho:p> 0.8 Ho: 0.8 Ho:p 0.8 H₁:µ 0.8 H₁:p 0.8 The test is: right-tailed two-tailed left-tailed = Based on a sample of 300 people, 82% owned cats The p-value is: (to 2 decimals)Use a y-test to test the claim o = 0.48 at the a = 0.01 significance level using sample statistics s = 0.466 and n = 15. Assume the population is normally distributed. Identify the null and alternative hypotheses. YA. Ho: o? = 0.48 O B. Ho: o 20.48 H:o +0.48 H,: o 0.48 Identify the test statistic. (Round to two decimal places as needed.)
- n engineer has designed a valve that will regulate water pressure on an automobile engine. The valve was tested on 160 engines and the mean pressure was 7.7 pounds/square inch (psi). Assume the population variance is 0.36. If the valve was designed to produce a mean pressure of 7.8 psi, is there sufficient evidence at the 0.01 level that the valve performs below the specifications? Step 2 of 3: Find the P-value for the hypothesis test. Round your answer to four decimal places.Test the claim that the proportion of men who own cats is significantly different than 90% at the 0.05 significance level.The null and alternative hypothesis would be: H0:μ=0.9H0:μ=0.9H1:μ>0.9H1:μ>0.9 H0:p=0.9H0:p=0.9H1:p>0.9H1:p>0.9 H0:p=0.9H0:p=0.9H1:p≠0.9H1:p≠0.9 H0:μ=0.9H0:μ=0.9H1:μ<0.9H1:μ<0.9 H0:p=0.9H0:p=0.9H1:p<0.9H1:p<0.9 H0:μ=0.9H0:μ=0.9H1:μ≠0.9H1:μ≠0.9 The test is: two-tailed right-tailed left-tailed Based on a sample of 25 people, 93% owned catsThe test statistic is: (to 2 decimals)The positive critical value is: (to 2 decimals)Based on this we: Fail to reject the null hypothesis Reject the null hypothesis Submit QuestionQuestion 12Using ANOVA, a null hypothesis could look like this: Group of answer choices H0: µ1 =µ2 =µ3 H0: µ1 > µ2 > µ3 H0: µ1 = µ2 > µ3 H0: µ1 ≠ µ2 ≠ µ3