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Question 853:



No answer provided yet.We want to compare the average number of take-out dinners between men and women and can compare means using the 2-sample z-test. We almost never know the population standard deviations and instead have to estimate them, we'd then use the t-test. In this example we've been given the two populations standard deviations so we can use the normal curve to answer our questions. The only thing we don't know are the population means, so we'll use the 2-sample z-test.

The null hypothesis is that there is no difference between the means. The alternative hypothesis is that there is a difference between means. We will reject the Null Hypothesis if our p-value is less than the alpha cut-off of .01.

The test statistic z* is found as the difference between the two means : 1.82.  We divide this difference by the standard error of difference by average the variances together based on the sample size. The variances are the square of the standard deviations. First divide each variance by the sample size, add them together, then take the square root = SQRT [20/35 + 14.9/40 ]. We get a standard error of the difference as .97.  Dividing this into the 1.82 difference gets us a test statistic of 1.87.

To find the p-value, we just need a normal table or can use the excel function =2*(1-NORMSDIST(1.87)) and we get the 2-tailed p-value of .061. Since this is greater than our alpha cut-off, we fail to reject the Null. We conclude there is not enough evidence to say there is a difference in mean take-outs times between men and women.

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