find out the critical value for two tailed test sample size = 39 test statistics = 1.365 level of significance = .10
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find out the critical value for two tailed test
test statistics = 1.365
level of significance = .10
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- poll used a sample of 100 randomly selected car owners. Within the sample, the mean time of ownership for a single car was 7.18 years with a standard deviation of 3.37 years. Test the claim by the owner of a large dealership that the mean time of ownership for all cars is less than 7.5 years. Use a 0.05 significance level. Find the Z score and P value and if left-tailed or right-tailed. Also to reject or not too.Is the Diet Practical? When 40 people used the Weight Watchers diet for one year, their mean weight loss was 3.0 lb and the standard deviation was 4.9 lb (based on data from “Comparison of the Atkins, Ornish, Weight Watchers, and Zone Diets for Weight Loss and Heart Disease Reduction, ” by Dansinger et al., Journal of the American Medical Association, Vol. 293, No. 1). Use a 0.01 significance level to test the claim that the mean weight loss is greater than 0. Based on these results, does the diet appear to be effective? Does the diet appear to have practical significance?A computer software vendor claims that a new version of its operating system will crash fewer than six times per year on average. A system administrator installs the operating system and claims that the new operating system will crash more than six times per year on average. At the end of year, the system administrator samples 41 computers with the new operating system and finds that the mean number of crashes on those computers for the year was 7.1 with sample standard deviation of 3.6 crashes. Can you conclude that the system administrator’s claim is correct? Use the critical value method to perform a hypothesis test at the 0.05 level of significance. a. Define the parameter we are testing and setup a hypothesis test to test the administrator’s claim: b. Calculate the test statistics for this hypothesis test. c. What is the critical value for this hypothesis test?
- In a test of weight loss programs, 111 subjects were divided such that 37 subjects followed each of 3 diets. Each was weighed a year after starting the diet and the results are in the ANOVA table below. Use a 0.05 significance level to test the claim that the mean weight loss is the same for the different diets. Source of Variation Between Groups Within Groups Total P-value 25.38918 0.8127 0.446353 F crit 3.080387 df MS F 50.778 3373.977 3424.755 2 108 31.24053 110 Should the null hypothesis that all the diets have the same mean weight loss be rejected? O A. Yes, because the P-value is greater than the significance level. O B. No, because the P-value is greater than the significance level. OC. No, because the P-value is less than the significance level. O D. Yes, because the P-value is less than the significance level.What is the estimated standard error for a paired sample t test using the following data: Standard deviation of the difference score = 5 and n=100?What is the value of the test statistic for a test with a null hypothesis that mu = 83, Xbar = 85, the sample standard deviation = 2 and n = 49?
- It is known that the amount of glass households in Phoenix recycled 10 years ago is normally distributed with a mean of 3.6 pounds of glass per week. Phoenix municipal waste recently chose a simple random survey of 24 households and found they recycled an average of 3.8 pounds a week with a standard deviation of 0.5 pounds. Using a significance level of 0.05, is this evidence that the amount of glass recycling has changed? Write a one or two sentence summary in non-statistical language to describe what you have found, not how you found it. Your audience is people who have never had a statistics class (think of middle of the road 8th graders, not the really bright ones). Is there anything about the sample data suggesting that the methods we learned should not be applied?What is the typical age of a person who renews his orher driver’s license online? From a random sample of660 driver’s license renewal transactions, the mean agewas 52.6 and the standard deviation was 6.4. Computethe 98% confidence interval estimate of the mean ageof online renewal users in this county.What is the value of the test statistic for a test with a null hypothesis that mu = 82, Xbar = 85, the sample standard deviation = 20 and n = 49?
