A clinical trial was conducted to test the effectiveness of the drug zopiclone for treating insomnia in older subjects. Before treatment with zopiclone, 16 subjects had a
- State the null hypothesis (H0).
- State the alternative hypothesis (H1).
- What kind of test would this be, a two-tailed, a left-tailed, or a right-tailed?
- Sketch a normal distribution curve and highlight the region of significance (also called critical region or rejection region).
- What test statistic is most relevant to this test? Why?\
- Use Excel to determine the P-value. What is it?
- On your graph, next highlight the P-value.
- State a conclusion about the null hypothesis. (Reject H0 or fail to reject H0.)
Next, test this hypothesis using the critical value method. - Using the most relevant test statistic, find the critical value from the tables at the back of the textbook. What is it?
- Sketch a normal distribution curve, plot the critical value, and highlight the critical or rejection region.
- Using the formulas in section 8-3, calculate the test statistic based on the sample data. Show your work. What is the test statistic?
- On your graph, plot the test statistic.
- State a conclusion about the null hypothesis. (Reject H0 or fail to reject H0.)
- Does zopiclone appear to be effective? Why do you think so?
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- A clinical trial was conducted to test the effectiveness of the drug zopiclone for treating insomnia in older subjects. Before treatment with zopiclone, 16 subjects had a mean wake time of 102.8 min. After treatment with zopiclone, the 16 subjects had a mean wake time of 98.9 min and a sample standard deviation of 42.3 min (based on data from “Cognitive Behavioral Therapy vs Zopiclone for Treatment of Chronic Primary Insomnia in Older Adults,” by Sivertsen et al., Journal of the American Medical Association, Vol. 295, No. 24). Assume that the 16 sample values appear to be from a normally distributed population, and test the claim that after treatment with zopiclone, subjects have a mean wake time of less than 102.8 min. 1. State the alternative and null hypothesis. 2. Is this a two tailed, left tailed or right tailed?arrow_forwardA pharmaceutical company needs to know if its new cholesterol drug, Praxor, is effective at lowering cholesterol levels. It believes that people who take Praxor will average a greater decrease in cholesterol level than people taking a placebo. After the experiment is complete, the researchers find that the 4848 participants in the treatment group lowered their cholesterol levels by a mean of 21.521.5 points with a standard deviation of 2.52.5 points. The 4040 participants in the control group lowered their cholesterol levels by a mean of 20.920.9 points with a standard deviation of 4.14.1 points. Assume that the population variances are not equal and test the company’s claim at the 0.100.10 level. Let the treatment group be Population 1 and let the control group be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to three decimal places. step 3 of 3 : Conclusion- fail to reject/reject...insufficient/sufficientarrow_forwardA study was conducted in which mice that had a dim light on at night (rather than complete darkness) ate most of their calories when they should have been resting. These mice gained a significant amount of weight, despite eating the same number of calories as mice kept in total darkness. The time of eating seemed to have a significant effect. There were 10 mice in the group with dim light at night and they gained an average of 7.9 g with a standard deviation of 3.0. We see in Figure 1 that the data are not heavily skewed and do not have extreme outliers. 00 O O 2 4 8 10 12 14 BM Gain Figure 1 Body mass gain (in grams) for mice with a night light Use the t-distribution to find a 90% confidence interval for weight gain. Round your answers to two decimal places. The 90% confidence interval is i to i !arrow_forward
- A technician compares repair costs for two types of microwave ovens (type I and type II). He believes that the repair cost for type I ovens is greater than the repair cost for type Il ovens. A sample of 60 type I ovens has a mean repair cost of $74.06, with a standard deviation of $16.10. A sample of 68 type Il ovens has a mean repair cost of $67.73, with a standard deviation of $14.70. Conduct a hypothesis test of the technician's claim at the 0.1 level of significance. Let u, be the true mean repair cost for type I ovens and u, be the true mean repair cost for type Il ovens. Step 1 of 4: State the null and alternative hypotheses for the test. Answer O Tables E Keypad Keyboard Shortcuts Submit Answer O 2020 Hawkes Learning E P Search for anything 812 AM 11/25/2020 hp brt se delete home end pg up 共 %24 4 2. 3 6 9 +] 8 + backspace num lock Q E R T Y 8. home G H K enter 4 pause V M t shift 2 end alt talt ctri insarrow_forwardA pharmaceutical company claims that its new drug is effective in reducing blood pressure with a mean reduction of at least 5 mmHg. To test this claim, a random sample of 30 patients is selected and given the medication. The results show an average reduction in blood pressure of 4.7 mmHg with a standard deviation of 1.2 mmHg. Test the company's claim at a 5% significance level.arrow_forwardA researcher is interested in examining the effect of emotions on those that are convicted of a crime while under the age of 15. She decides to compare the empathy quotient (EQ) scores of those convicted of violent crimes to those convicted of non-violent crimes. A random sample of 35 violent offenders had a mean score of 33.5 and a standard deviation of 3.2, and another sample of 46 non-violent offenders had a mean of 37 and a standard deviation of 3.8. Also, suppose that equal variance cannot be assumed, but the populations are normal. If a higher EQ score indicated increased empathy, construct a 95% confidence interval for the difference in means. What is the critical value? What is the lower bound of the confidence interval? What can she conclude about the difference in empathy between the two groups?arrow_forward
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