You test the following hypotheses: Ho: µ = 1000 Ha: µ + 100O You calculate a test statistic z = -3.13. You should reject the null hypothesis at a 99% confidence level (alpha = .01). O True O False
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A: Before we proceed with the hypothesis test, we must calculate necessary descriptive statistics…
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A: GIVEN DATA, n=374x=45p^=xn=45374=0.1203claim:p≠0.18
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A: Note: Hey there! Thank you for the question. As you have posted a question with several sub-parts,…
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A: Sample size n =16Standard deviation s=74
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Q: Only about 15% of all people can wiggle their ears. Is this percent different for millionaires? Of…
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Q: Only about 18% of all people can wiggle their ears. Is this percent different for millionaires? Of…
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Q: Only about 16% of all people can wiggle their ears. Is this percent lower for millionaires? Of the…
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Q: 9.3.5 x Your answer incorrect. Try again. For the hypothesis test Ho : H = 24 against H :p < 24 with…
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Q: 18% of all college students volunteer their time. Is the percentage of college students who are…
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A: Here we don't know the population standard deviation. We use t test for one mean.
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A: From the provided information, The hypotheses are as follow: H0: µ = 3.54 Ha: µ < 3.54
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Q: A study was conducted to determine whether magnets were effective in treating pain. The values…
A: n1 = 20n2 = 20x¯1 = 0.43x¯1 = 0.47s1 = 1.37s2 = 0.94 α = 0.05claim : σ12 > σ22
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Q: Hello. I appreciate any help. Thanks.
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A: Givenn=326x=49α=0.01p^=xn=49326=0.1503
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- A study was conducted to determine whether magnets were effective in treating pain. The values represent measurements of pain using the visual analog scale. Assume that both samples are independent simple random samples from populations having normal distributions. Use a 0.05 significance level to test the claim that those given a sham treatment have pain reductions that vary more than the pain reductions for those treated with magnets. What are the null and alternative hypotheses? OA. Ho: 002 H₁:0² = 0²/2 OC. H₂:0²=0²2 H₁:0² > 0²/2 Identify the test statistic. F = (Round to two decimal places as needed.) Use technology to identify the P-value. The P-value is (Round to three decimal places as needed.) What is the conclusion for this hypothesis test? C OB. H₁: 0 = 0² H₁:0² +03 OD. H₂:0²=0²2 H₁:0² <0²/ O A. Reject Ho. There is sufficient evidence to support the claim that those given a sham treatment have pain reductions that vary more than those treated with magnets. O B. Fail to reject…A random sample of 60 trials resulted in 18 successes. List the null and alternate hypothesis if you were to test the claim that the population proportion exceeds 20% at an ?=0.01.20% of all college students volunteer their time. Is the percentage of college students who are volunteers smaller for students receiving financial aid? Of the 356 randomly selected students who receive financial aid, 61 of them volunteered their time. What can be concluded at the αα = 0.10 level of significance? For this study, we should use? Select an answer z-test for a population proportion? or t-test for a population mean? The null and alternative hypotheses would be: H0:H0:? μ or p Select an answer ≥ = < > ≤ ≠ _______(please enter a decimal) H1:H1:? μ or p Select an answer ≥ ≠ = < > ≤ _______ (Please enter a decimal) 2. The test statistic ? t or z = ______ (please show your answer to 3 decimal places.) The p-value = ______ (Please show your answer to 4 decimal places.) The p-value is ? > or ≤ α 3. Based on this, we should Select an answer fail to reject? reject ? accept ? the null…
- You wish to test the claim that the average IQ score is less than 100 at the .01 significance level. You determine the hypotheses are: Ho: μ=100 H1:μ<100H You take a simple random sample of 79 individuals and find the mean IQ score is 98.3, with a standard deviation of 15.3. Let's consider testing this hypothesis two ways: once with assuming the population standard deviation is not known and once with assuming that it is known. Round to three decimal places where appropriate. Assume Population Standard Deviation is NOT known Assume Population Standard Deviation is 15 Test Statistic: t = Test Statistic: z = Critical Value: t = Critical Value: z = p-value: p-value: Conclusion About the Null: Reject the null hypothesis Fail to reject the null hypothesis Conclusion About the Null: Reject the null hypothesis Fail to reject the null hypothesis Conclusion About the Claim: There is sufficient evidence to support the claim that the average IQ score is less…After running a regression you conduct a hypothesis test of Ho: ß = 2.5 versus Hạ: B + 2.5 is performed using a = 0.10. The value of the test statistic z = -1.80 (the critical value is -1.645). If the true value of B is 2.5, does the conclusion you reach result in Type I, Type II error, or the correct decision? O Type II Error O Type I Error O correct decision not enough informationSuppose that you are testing the following hypotheses: Ho: p = .80 and Ha: p # .80. You calculate a test statistic value of z = -6.08. True or False: At a 95% confidence level, which is alpha = .05, you should reject the null hypothesis. True False
- Please answer question 10.5In their advertisements, a new diet program would like to claim that their methods result in a mean weight loss of more than ten pounds in two weeks. In order to determine if this is a valid claim, they hire an independent testing agency that then selects twenty-five people to be placed on this diet. Which of the following is the correct hypotheses? a Ho: μ- 10 Ha: μ > 10 Ho: u > 10 Ha: u = 10 Ho: U = 10 Ha: u < 10 O d Ho: p = 10 Ha: u+ 10You wish to test the claim that the average IQ score is less than 100 at the .10 significance level. You determine the hypotheses are: Ho: μ=100Ho: μ=100 H1:μ<100H1:μ<100 You take a simple random sample of 73 individuals and find the mean IQ score is 95.2, with a standard deviation of 15.1. Let's consider testing this hypothesis two ways: once with assuming the population standard deviation is not known and once with assuming that it is known. Round to three decimal places where appropriate. Assume Population Standard Deviation is NOT known Assume Population Standard Deviation is 15 Test Statistic: t = Test Statistic: z = Critical Value: t = Critical Value: z = p-value: p-value: Conclusion About the Null: Reject the null hypothesis Fail to reject the null hypothesis Conclusion About the Null: Reject the null hypothesis Fail to reject the null hypothesis Conclusion About the Claim: There is sufficient evidence to support the claim that the average IQ…
- Under what circumstances is a t statistic used instead of a z-score for a hypothesis test? Justin wants to know whether a commonly prescribed drug does improve the attention span of students with attention deficit disorder (ADD). He knows that the mean attention span for students with ADD who are not taking the drug is 2.3 minutes long. His sample of 12 students taking the drug yielded a mean of 4.6 minutes. Justin can find no information regarding σx , so he calculated s2x =1.96. Determine the critical region using a one-tailed test with alpha = .05. Conduct the hypothesis test (Do the math and compare the t-critical and t-obtained values). State your conclusions in terms of H0 (Should you reject the H0 or fail to reject/accept the H0). Based on your analysis, is there a relationship between the drug and attention span?A random sample of n = 25 is obtained from a population with variance ?2, and the sample mean is computed to be ?̅= 70. Consider the null hypothesis ?0: ? =80 versus the alternative hypothesis ?1: ? < 80. Compute the p-value when the population variance is ?2 = 600.Fil 10. A small pilot study is conducted to investigate the effect of a nutritional supplement on total body weight. Six participants agree to take the nutritional supplement. To assess its effect on body weight, weights are measured before starting the supplementation and then after 6 weeks. The data are shown below. Is there a significant increase in body weight following supplementation? Run the test at a 5% level of significance. Demonstrate the five steps for hypothesis testing and explain your results. Subject İnitial Weight Weight after 6 Weeks 1 155 157 2 142 145 3 176 180 4 180 175 210 209 6. 125 126