A simple random sample of 16 subjects with the treatment of zopiclone 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 standard deviation of 42.3 min. Use a 0.05 significance level to claim that after treatment with zopiclone, subjects had a mean waketime less than 102.8 min.
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- A scale measuring confidence in the media was administered to a random sample of politically-affiliated Americans, with higher scores indicating greater levels of confidence. A sample of 26 Democrats had a mean score of 8.2 on the scale with a standard deviation of 1.5. A sample of 24 Republicans scored an average of 7.8 with a standard deviation of 1.1. Is there a significant difference between Democrats and Republicans regarding confidence in the media? Conduct a significance test with alpha = .05. STEP ONE: What is the fourth requirement for this significance test? 1. Independent random samples 2. Interval-ratio variable of confidence in the media (technically ordinal) 3. Normal sampling distribution assumed 4.A sample of 200 corporate managers produced the mean job-related stress score of 7.9 with a standard deviation of .65. Another sample of 300 college professors produced the mean job-related stress score of 5.4 with a standard deviation of .90. The null hypothesis is that the mean job-related stress score for all corporate managers is equal to the mean job-related stress score for all college professors. The alternative hypothesis is that the mean job-related stress score for all corporate managers is not equal to the mean job-related stress score for all college professors. The significance level is 1%. What are the critical values of z? a. –2.33 and 2.33 b. –2.17 and 2.17 c. –1.96 and 1.96 d. –2.58 and 2.58Volunteers who had developed a cold within the previous 24 hours were randomized to take either zinc or placebo lozenges every 2 to 3 hours until their cold symptoms were gone. Twenty-five participants took zinc lozenges, and 23 participants took placebo lozenges. The mean overall duration of symptoms for the zinc lozenge group was 4.5 days, and the standard deviation of overall duration of symptoms was 1.6 days. For the placebo group, the mean overall duration of symptoms was 8 days, and the standard deviation was 1.8 days. (a) Calculate a 95% confidence interval for the mean overall duration of symptoms in a population of individuals like those who used the zinc lozenges. (Round your answers to two decimal places.)to days(b) Calculate a 95% confidence interval for the mean overall duration of symptoms in a population of individuals like those who used the placebo lozenges. (Round your answers to two decimal places.)to days(c) On the basis of the intervals computed in parts (a) and…
- In a test of the effectiveness of garlic for lowering cholesterol, 36 subjects were treated with raw garlic Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.1 and a standard deviation of 23.2. Use a 0.01 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. What are the null and alternative hypotheses? O A. Ho p=0 mg/dL O B. H, p=0mg/dL H, p0 mg/dL OC. Ho H>0 mg/dL O D. H, p=0 mg/dL. H, pz0 mg/dL. H, p<0 mg/dLA new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 10 days following no advertisements, the mean was 18.3 purchasing customers with a standard deviation of 1.8 customers. On 7 days following advertising, the mean was 19.4 purchasing customers with a standard deviation of 1.6 customers. Test the claim, at the 0.02 level, that the mean number of customers who make a purchase in the store is lower for days following no advertising compared to days following advertising. Assume that both populations are approximately normal and that the population variances are equal. Let days following no advertisements be Population 1 and let days following advertising be Population 2. Step 3 of 3: Draw a…Bone mineral density (BMD) is a measure of bone strength. Studies show that BMD declines after age 45. The impact of exercise may increase BMD. A random sample of 59 women between the ages of 41 and 45 with no major health problems were studied. The women were classified into one of two groups based upon their level of exercise activity: walking women and sedentary women. The 39 women who walked regularly had a mean BMD of 5.96 with a standard deviation of 1.22. The 20 women who are sedentary had a mean BMD of 4.41 with a standard deviation of 1.02. Which of the following inference procedures could be used to estimate the difference in the mean BMD for these two types of women
- A car company claims that its new SUV gets better gas mileage than its competitor's SUV. A random sample of 49 of its SUVs has a mean gas mileage of 17.1 miles per gallon (mpg). The population standard deviation is known to be 1.3 mpg. A random sample of 33 competitor's SUVS has a mean gas mileage of 16.3 mpg. The population standard deviation for the competitor is known to be 1.6 mpg. Test the company's claim at the 0.05 level of significance. Let the car company be Population 1 and let the competitor be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to two decimal places.Dr. Graham is interested in determining if middle-aged adults use text messaging more or less frequently than the general population. Dr. Graham collects information on text messaging from a random sample of 50 adults ages 25 to 44. Dr. Graham finds that these individuals send or receive an average of 68 text messages per day. Using the population mean (and standard deviation) of 41.5 texts per day (34 texts per day), determine whether adults in this age group use text messaging more than the general public.A new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 7 days following no advertisements, the mean was 22.1 purchasing customers with a standard deviation of 1.2 customers. On 10 days following advertising, the mean was 24.1 purchasing customers with a standard deviation of 1.6 customers. Test the claim, at the 0.05 level, that the mean number of customers who make a purchase in the store is lower for days following no advertising compared to days following advertising. Assume that both populations are approximately normal and that the population variances are equal. Let days following no advertisements be Population 1 and let days following advertising be Population 2. Step 1 of 3: State the…
- In a test of the effectiveness of garlic for lowering cholesterol, 4949 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.30.3 and a standard deviation of 2.142.14 Use a 0.100.10 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 00. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. What are the null and alternative hypotheses? A. Upper H 0H0: muμequals=00 mg/dL Upper H 1H1: muμless than<00 mg/dL B. Upper H 0H0: muμequals=00 mg/dL Upper H 1H1: muμgreater than>00 mg/dL C. Upper H 0H0: muμequals=00 mg/dL Upper…A new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 13 days following no advertisements, the mean was 23.9 purchasing customers with a standard deviation of 1.9 customers. On 6 days following advertising, the mean was 24.7 purchasing customers with a standard deviation of 1.6 customers. Test the claim, at the 0.01 level, that the mean number of customers who make a purchase in the store is lower for days following no advertising compared to days following advertising. Assume that both populations are approximately normal and that the population variances are equal. Let days following no advertisements be Population 1 and let days following advertising be Population 2. Step 3 of 3: Draw a…In a test of the effectiveness of garlic for lowering cholesterol, 4949 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.40.4 and a standard deviation of 23.823.8. Use a 0.010.01 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 00. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. Determine the test statistic. nothing (Round to two decimal places as needed.)