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1. If Ho is rejected at 0.05 level of significance, then Ho will also be rejected at
0.10 level of significance using the same test on the same dataset.
2. A 95% confidence
hypothesis Ho: μ = 20 vs Ha: μ ≠ 20 at 0.05 level of significance.
3. In testing for a disease, the null hypothesis is the person is healthy while the
alternative hypothesis is the person has the disease. If a healthy person is
diagnosed as having the disease, a Type I error is committed.
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- In a test of the effectiveness of garlic for lowering cholesterol, 47 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before−after) in their levels of LDL cholesterol (in mg/dL) have a mean of 2.8 and a standard deviation of 16.3. Construct a 90% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol?Scores on a certain "IQ" test for 18-25 year olds are normally distributed. A researcher believes that the average IQ score for students at a certain NJ college is less than 110 points, and so wants to test this hypothesis. The researcher obtain a SRS of 45 student IQ scores from school records and found the mean of the 45 results was 108 with a sample standard deviation of 21. The level of significance (alpha) used for this problem is 0.05. What is the appropriate test statistic (Student must complete by showing by formula using the ap- propriate values in that formula "showing work" and the final answer and appropriate label)? O T test score = (108-110)/(21/sqrt(45)) = -.639 T test score (108-110)/(21/sqrt(45)) = .639 %3D OT test score = (110-108)/(21/sqrt(45)) = .639 %3D T test score (108-110)/(45/sqrt(21)) =-.2037 %3D 素The Woodcock-Johnson Cognitive Ability test indicates that the average score is 95. The same group of 12th grade students (n = 72) in Columbus are tested and the sample mean is 110.1 and the sample standard deviation is 15.2. Do the scores provide good evidence that the mean IQ of this population is greater than 95? State the null (H0) and alternative (Ha) hypotheses. What are the confidence intervals for this data? What is the value of the test statistic? What is the p-value for this data? What is your decision about the null hypothesis? summerize the conclusion in the context
- Does the average Presbyterian donate less than the average Catholic in church on Sundays? The 58 randomly observed members of the Presbyterian church donated an average of $24 with a standard deviation of $15. The 42 randomly observed members of the Catholic church donated an average of $31 with a standard deviation of $7. What can be concluded at the αα = 0.10 level of significance? For this study, we should use Select an answer t-test for the difference between two dependent population means z-test for a population proportion t-test for a population mean t-test for the difference between two independent population means z-test for the difference between two population proportions The null and alternative hypotheses would be: H0:H0: Select an answer p1 μ1 Select an answer ≠ > = < Select an answer p2 μ2 (please enter a decimal) H1:H1: Select an answer p1 μ1 Select an answer < = ≠ > Select an answer p2 μ2 (Please enter a decimal) The test statistic ?…Suppose one would like to contract a 99.9% confidence interval for a sample mean when the population standard deviation is known. Determine the Z value that would be used to contract such a confidence interval run the solution to two decimal places. Z=Develop a 95% confidence interval for the population slope if the followingregression information are given: b1= 41.7, p-value = 0.02, and n = 35.
- A survey found that the average daily cost to rent a car in Los Angeles is $102.24 and in Las Vegas is $97.35. The data were collected from two random samples of 40 in each of the two cities and the population standard deviations are $5.98 for Los Angeles and $4.21 for Las Vegas. At the 0.05 level of significance, construct a confidence interval for the difference in the means and then decide if there is a significant difference in the rates between the two cities. Let the sample from Los Angeles be Group 1 and the sample from Las Vegas be Group 2. Confidence Interval (round to 4 decimal places)In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before−after) in their levels of LDL cholesterol (in mg/dL) have a mean of 5.3 and a standard deviation of 17.4. Construct a 90% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol? . What is the confidence interval estimate of the population mean μ? enter your response here mg/dL<μ<enter your response here mg/dL (Round to two decimal places as needed.)A soap bar produced by a company is claimed to have a mean net weight of 4.5 ounces. A consumer agency wants to test this claim. A sample of 80 soap bars of this brand gave a mean net weight of 4.46 ounces with a standard deviation of .06 ounces. Test at the 1% significance level if the mean net weight of these soap bars is less than 4.5 ounces. 1. Null hypothesis: S [ Select ] 2. Alternative hypothesis: [ Select ] 3. Critical values: [Select] 4. Test Statistic: [ Select ] 5. Conclusion: [Select ] 6. Summary statement: I [ Select ]
- A data set includes data from 500 random tornadoes. The display from technology available below results from using the tornado lengths (miles) to test the claim that the mean tornado length is greater than 2.2 miles. Use a 0.05 significance level. Use the display to identify the null and alternative hypotheses, test statistic, and P-value. State the final conclusion that addresses the original claim.In a test of the effectiveness of garlic for lowering cholesterol, 45 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before−after)in their levels of LDL cholesterol (in mg/dL) have a mean of 4.6 and a standard deviation of 16.3. Construct a 95% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol? What is the confidence interval estimate of the population mean μ? ______ mg/dL<μ<_____ mg/dL (Round to two decimal places as needed.) What does the confidence interval suggest about the effectiveness of the treatment? A. The confidence interval limits contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels. B. The confidence interval limits contain 0, suggesting that the garlic treatment…Preterm infants with late metabolic acidosis (n=55) were compared to preterm infants without the condition (n=70) with respect to the level of a certain chemical in their urine. The infants with the condition had a mean level of 8.9 with a standard deviation of 4.3, while infants without the condition had a mean level of 5.0 with a standard deviation of 3.8. Construct a two-sided 95% confidence interval for the difference between the two means. Assume equal variances. You MUST show your work to receive full credit. Partial credit is available. Give the following in your answer: 1. The value for t and how you found it, 2. The calculation of the pooled variance, 3. The calculation of the Standard Error, 4. The calculation of the Margin of Error, 5. The calculation of the upper and lower limits, and 6. An interpretation of the confidence interval. Upload an image, pdf, or Word file with pictures of your handwritten work.
