Find a 99% confidence interval in the difference of the means. Use technology to find the critical value using df = 44.7323 and round answers to 4 decimal places. < μ1-μ2 <
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A school administrator wants to see if there is a difference in the number of students per class for Portland Public School district (group 1) compared to the Beaverton School district (group 2). Assume the populations are
Find a 99% confidence interval in the difference of the means. Use technology to find the critical value using df = 44.7323 and round answers to 4 decimal places.
< μ1-μ2 <
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- The National Coalition on Healthcare suggests that the mean annual premium that a health insurer charges an employer for a health plan covering a family of four averaged $13,000 in 2009. A sample of 30 families of four yields a mean annual premium paid by their employer to be $13,500 with a sample standard deviation of $300. We are interested in whether the mean annual premium that a health insurer charges an employer for a health plan covering a family of four is different from $13,000 using a significance level of 0.10. Select one: a. tdata = 9.129, do not reject Ho. b. tdata = 9.129, reject Ho. c. tdata = -9.129, do not reject Ho. O d. tdata = 4.156, reject Ho.A hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 7.1 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 6.1, 8.1, 9.1, 10.1, 7.1, 11.1, and 12.1 pounds. What is the sample standard deviation?Students preparing for comprehensive exams usually spend many hours studying. Suppose it is known that the time students spend studying for comprehensive exams has a distribution that is skewed heavily to the right with a mean of 41.3 hours and a standard deviation of 6.8 hours. If a simple random sample of 72 students is selected and the amount of time each spent studying for the comprehensive exam is determined. State and check the two assumptions.
- Fran is training for her first marathon, and she wants to know if there is a significant difference between the mean number of miles run each week by group runners and individual runners who are training for marathons. She interviews 42 randomly selected people who train in groups and finds that they run a mean of 47.1 miles per week. Assume that the population standard deviation for group runners is known to be 4.4 miles per week. She also interviews a random sample of 47 people who train on their own and finds that they run a mean of 48.5 miles per week. Assume that the population standard deviation for people who run by themselves is 1.8 miles per week. Test the claim at the 0.01 level of significance. Let group runners training for marathons be Population 1 and let individual runners training for marathons be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.A sample of 76 female workers and another sample of 48 male workers from a state produced mean weekly earnings of $743.50 for the females and $777.63 for the males. Suppose that the population standard deviations of the weekly earnings are $80.05 for the females and $88.68 for the males. The null hypothesis is that the mean weekly earnings are the same for females and males, while the alternative hypothesis is that the mean weekly earnings for females is less than the mean weekly earnings for males. Directions: • Label your answers with the correct statistical symbols. • If you use the Ti, identify which function and values you used to calculate. If you solve by hand, show all steps 2.5 The significance level for the test is 1%. What is/are the critical value(s)? 2.6. What is the value of the test statistic, rounded to three decimal places? 2.7. What is the p-value for this test, rounded to four decimal places? 2. 8. Using the p-value approach, do you reject or fail to reject the null…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 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…
- Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 12people who buy insurance from Company A, the mean cost is $153 per month with a standard deviation of $16. For 15 randomly selected customers of Company B, you find that they pay a mean of $160 per month with a standard deviation of $10. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.10 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. H0: μ1=μ2 Ha: μ1_____μ2 Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places Step 3 of…Researchers want to compare the relative effectiveness of two popular diets. They randomly assign each of 400 volunteers to one of two Groups: A & B. There are 200 volunteers in each group. Group A spends 6 months on Diet A. Group B spends 6 months on Diet B. At the beginning of the study, the difference in average weight between the two groups was negligible. After the study, the Group A had lost on average 7.8 pounds with a standard deviation of 13.4 pounds, while the Group B had lost on average 5.3 pounds with a standard deviation of 14.8 pounds. Determine the z-score/test statistic for a two-tailed test with significance level α = 0.05 , accurate to 2 decimal places.An engineer designed a valve that will regulate water pressure on an automobile engine. The engineer designed the valve such that it would produce a mean pressure of 5.5 pounds/square inch. It is believed that the valve performs above the specifications. The valve was tested on 24 engines and the mean pressure was 5.7 pounds/square inch with a variance of 0.49. A level of significance of 0.01 will be used. Assume the population distribution is approximately normal. Make the decision to reject or fail to reject the null hypothesis.
- According to Kaiser Family Foundation survey in 2011 and 2010, the average premium for employer-sponsored health insurance for family coverage was $15,073 in 2011 and $13,770 in 2010 (USA TODAY, September 29, 2011). Suppose that these averages were based on random samples of 250 and 200 employees who had such employer-sponsored health insurance plans for 2011 and 2010, respectively. Further assume that the population standard deviations for 2011 and 2010 were $2160 and $1990, respectively.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…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…
