(a) 95% confidence interval for p: 0.07 to 0.12 Conclusion: Significance level: Sign of the correlation: (b) 90% confidence interval for p: -0.12 to -0.01 Conclusion: Significance level: Sign of the correlation: Conclusion: Ho (c) 99% confidence interval for p: -0.02 to 0.04 Significance level: Sign of the correlation: Ho Ho
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- Researchers interested in determining the relative effectiveness of two different drug treatments on people with a chronic illness established two independent test groups. The first group consisted of 12 people with the illness, and the second group consisted of 14 people with the illness. The first group received treatment 1 and had a mean time until remission of 166 days with a standard deviation of 8 days. The second group received treatment 2 and had a mean time until remission of 163 days with a standard deviation of 9 days. Assume that the populations of times until remission for each of the two treatments are normally distributed with equal variance. Construct a 90% confidence interval for the difference −μ1μ2 between the mean number of days before remission after treatment 1 ( μ1 ) and the mean number of days before remission after treatment 2 ( μ2 ). Then find the lower limit and upper limit of the 90% confidence interval. Carry your intermediate…Two catalysts in a batch chemical process are being compared for their effect on the output of the process reaction. A sample of 14 batches was prepared using catalyst 1 and gave an average yield of 85 with a sample standard deviation of 4. A sample of 13 batches was prepared using catalyst 2 and gave an average yield of 80 and a sample standard deviation of 3. Find a 90% confidence interval for the difference between the population means, assuming that the populations are approximately normally distributed with equal variances.1. A sample of 205 body temperatures with a mean of 98.49oF and a standard deviation of 0.58oF. Use a 0.01 significance level to test the claim that the mean body temperatures of the population is equal to 98.6oF. Find the test statistic. Round to four decimal places. 2.A sample of 205 body temperatures with a mean of 98.49oF and a standard deviation of 0.58oF. Use a 0.01 significance level to test the claim that the mean body temperatures of the population is equal to 98.6oF. Find the p-value that associated with the test statistic. Round to four decimal places.
- Medical researchers interested in determining the relative effectiveness of two different drug treatments on people with a chronic mental illness established two independent test groups. The first group consisted of 10 people with the illness, and the second group consisted of 14 people with the illness. The first group received treatment 1 and had a mean time until remission of 185 days, with a standard deviation of 5 days. The second group received treatment 2 and had a mean time until remission of 179 days, with a standard deviation of 8 days. Assume that the populations of times until remission for each of the two treatments are normally distributed with equal variance. Can we conclude, at the 0.05 level of significance, that the mean number of days before remission after treatment 1, μ1 , is greater than the mean number of days before remission after treatment 2, μ2 ? Perform a one-tailed test. Then fill in the table below. Carry your intermediate…A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.61 hours, with a standard deviation of 2.39 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.43 hours, with a standard deviation of 1.61 hours. Construct and interpret a 90%confidence interval for the mean difference in leisure time between adults with no children and adults with children μ1−μ2. Let μ1 represent the mean leisure hours of adults with no children under the age of 18 and μ2 represent the mean leisure hours of adults with children under the age of 18. -The 90% confidence interval for μ1−μ2is the range from____ hours to____hours. (Round to two decimal places as needed.) -What is the interpretation of this confidence interval? A. There is 90% confidence that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference…Medical researchers interested in determining the relative effectiveness of two different drug treatments on people with a chronic mental illness established two independent test groups. The first group consisted of 8 people with the illness, and the second group consisted of 14 people with the illness. The first group received treatment 1 and had a mean time until remission of 163 days, with a standard deviation of 9 days. The second group received treatment 2 and had a mean time until remission of 174 days, with a standard deviation of 8 days. Assume that the populations of times until remission for each of the two treatments are normally distributed with equal variance. Can we conclude, at the 0.05 level of significance, that the mean number of days before remission after treatment 1, μ1, differs from the mean number of days before remission after treatment 2, μ2? Perform a two-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal…
- Seat Belt ProblemA random sample of car crashes is obtained. Among 2792 occupants not wearing seat belts, 33 were killed. Among 7654 occupants wearing seat belts, 18 were killed. Use a 0.01 significance level to test the claim that seat belts are effective in reducing fatalities. Consider the occupants not wearing seat belts to be population 1. State the final conclusion regarding the claim (using confidence interval). Because the confidence interval limits __________ 0, there __________ appear to be a significant difference between the two proportions. There __________ sufficient evidence to support the claim that seat belts are effective in reducing fatalities.a) Soil has been improved to increase the strength of a ground. The average compressive strength of 61 samples before treatment is 45 kg / cm2 and the standard deviation is 6.75 kg / cm2; The average compressive strength of 56 samples after the improvement is 52.5 kg / cm2 and the standard deviation is 8 kg / cm2. Determine whether the improvement application changes the standard deviation of the soil strength or not at the 5% significance level (15 P). b) Find the confidence intervals at the 7% significance level of the mean compressive strength of the sample after improvement?A random sample of adults were asked whether they prefer reading an e-book over a printed book. The survey resulted in a sample proportion of p^=0.14, with a sampling standard deviation of σp^=0.02, who preferred reading an e-book. Use the Empirical Rule to construct a 95% confidence interval for the true proportion of adults who prefer e-books.
- A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.03 hours, with a standard deviation of 2.46 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.48 hours, with a standard deviation of 1.52 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children μ1−μ2. Let μ1 represent the mean leisure hours of adults with no children under the age of 18 and μ2 represent the mean leisure hours of adults with children under the age of 18. The 90%confidence interval for μ1−μ2 is the range from nothinghours to nothing hours. Also interpret the confidence interval. (Round to two decimal places as needed.)A tire manufacturer claims that its tires have a mean life of at least 50,000 km. A random sample of 25 of these tires is tested and the mean life is 33,000 km. Suppose the population standard deviation is 2,500 km. and the lives of tires is approximately normal. To test the manufacturer’s claim at the 5% level of significance, the analyst should A. perform a right-tailed test using the t-statistic.B. perform a left-tailed test using the t-statistic.C. perform a right-tailed test using the z- statistic.D. perform a left-tailed test using the z- statistic.A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.65 hours, with a standard deviation of 2.29 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.26 hours, with a standard deviation of 1.98 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children (μ1−μ2.) Let μ1 represent the mean leisure hours of adults with no children under the age of 18 and μ2 represent the mean leisure hours of adults with children under the age of 18. b. The 90% confidence interval for μ1−μ2 is the range from ____ hours to _____ hours. (Round to two decimal places as needed.) What is the interpretation of this confidence interval? A. There is 90% confidence that the difference of the means is in the interval. Conclude that there is insufficient evidence of a…