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Question 844:



No answer provided yet.We can use the 1-sample z-test to test the sample mean against the test criteria of 10lbs. The Null Hypothesis is that the reduction in weight loss is equal to or greater than 10lbs. The alternative hypothesis is that the mean weight loss is less than 10lbs. We will reject the Null Hypothesis if the p-value is less than the alpha cut-off value of .05.

The test statistic z is found as the difference between the sample mean and test mean (10-9) = 1lb. We divide this difference by the standard error of the mean (SE). The SE is found as the standard deviation divided by the square root of the sample size = 2.8/SQRT(50) = .3959. The test statistic z = 1/.3959 = 2.525.

The one-sided p-value associated with this z-score is .00578 which is far less than .05. The p-value can be found by looking up 2.525 in a normal table or using the excel function =NORMSDIST().

Since it is less than the alpha cutoff of .05 we reject the Null Hypothesis and conclude there is sufficient evidence to say the mean reduction is less than 10lbs.

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