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When 40 people used the Weight Watchers diet for one year, their mean weight loss was 3.00 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). Assume that the standard deviation of all such weight changes is ? = 4.9 lb. We shall use a 0.01 significance level to test the claim that the mean weight loss is greater than 0.
a) Set up the null and alternative hypotheses, and perform the hypothesis test.
b) Based on these results, does the diet appear to be effective? Does the diet appear to have practical significance?
Bicycle helmet use. Table lists data from a cross-sectional survey of bicycle safety. The explanatory variable is a measure of neighbourhood socioeconomic status (variable_RFM). The response variable is "percent of bicycle riders wearing a helmet"..
Is there evidence that teams with higher total payrolls tend to be more successful? Justify your answer.
For a new HDTV the mean number of months until repairs are needed is 36.84 with a standard deviation of 3.34 months.
Elucidate the null and alternative hypothesis for the difference among the two means. In testing the difference between the means of two.
Draw suitable distribution curve indicating position of the TS and CV.
Using the original data, test at the 5% significance level that the population standard deviation of studying times is greater than 4 hours.
Plot the data on the same line and give your impression regarding any apparent differences between the two companies.
Suppose that you are conducting a survey of customer satisfaction regarding their cellular phone. If you called 11 households selected at random, what is the probability that:
Construct a 95% confidence interval for the population mean.
Assume that for population of students at school, distribution of x is well approximated by normal curve with mean 48 min and standard deviation 4 min.
When we carry out a goodness of fit chi-square test, the expected frequencies are based on the alternative hypothesis
Use a test of the difference between proportions when answering these questions.
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