When isabel blancarte was hired as a waitress in the student...

Name: when isabel blancarte was hired as a waitress in the student cafeteria at the veracruz institute of professional studies she was told you can earn on average more than 200 a day in tips in t 33703
Uploaded: 2022-01-26T18:19:56-08:00
Duration: 5 min 27 s
Channel: Robin Corrigan
Description: when isabel blancarte was hired as a waitress in the student cafeteria at the veracruz institute of professional studies she was told you can earn on average more than 200 a day in tips in t 33703

When Isabel Blancarte was hired as a waitress in the student cafeteria at the Veracruz Institute of Professional Studies, she was told, “You can earn, on average, more than $200 a day in tips.” In the first 35 days of her employment at the cafeteria, the average daily amount of tips received was $248.50. At the .01 significance level, can Ms. Blancarte conclude that she is earning more than $200 in tips? Assume a population standard deviation of $32.40.

Transcript

00:01 In this question, we want to test whether a server can earn an average of more than $80 in tips per day.

00:08 And so the server has a sample of 35 workdays, and the average earned in this 35 days was $84 .85, with the sample standard deviation of 11 .38.

00:21 And so we are asked to test the claim at a significance level of 0 .01, whether the mean is actually greater than 80.

00:28 So for part a, we state our hypotheses.

00:35 So we can say that the null hypothesis is that the mean is no greater than 80.

00:46 And the alternative hypothesis then is that the mean is greater than 80.

00:55 And then for part b, we are asked to define our rejection region.

01:10 So two things we note here.

01:11 This is a one -tailed or right -tailed test.

01:14 And the significance level is .01.

01:19 So that means that we will reject the no hypothesis.

01:22 If our test statistic is greater than 2 .33, our 2 .36.

01:34 Next, for part c, we actually calculate our test statistic.

01:39 Now we can note that we don't have the population standard deviation, but our sample size is greater than 30, and so we can use the sample standard deviation instead of the population standard deviation, which means that we can use the following test statistic.

01:56 So the sample average is 84 .85, the null hypothesized name is 80, sample standard deviation 11 .38, and sample size is 35...

Question

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