Questions 6 - 9 are based on the following: The Director of...

Questions 6 - 9 are based on the following: The Director of Student Health Services wishes to know if her anti-smoking campaign has been effective. Last year, a complete survey of all students found the mean number of cigarettes smoked per day by students to be 8.6 with a standard deviation of 2.0. The Director of the Student Health Service obtains the following results from a recently selected 100 random sample of students (in cigarettes smoked per day): mean = 7.7. Perform the hypothesis testing with a one-tailed test at ??? = .05. What is the statistical conclusion? Reject the null hypothesis. Retain the null hypothesis.

Transcript

00:01 So our question says that the director of student health services wishes to know if our antispochamping has been affected.

00:07 Last year, a complete survey of all students found the main number of cigarettes smoked per day by students to be 8 .4 with a standard deviation of 2 .0.

00:16 So the director of the student health service obtained the following results from a recently selected random sample of 100 students.

00:23 We have the main to be across to 7 .7.

00:26 So we are supposed to perform the hypothesis test with a 1 -tile test at 5 % level of significance.

00:32 So let's go into our worksheet.

00:34 Our population mean m is equals to 8 .6.

00:37 Our population standard deviation zignor is 2 .0.

00:41 The sample size n is equals to 100.

00:45 And our sample mean x bar is equals to 7 .7.

00:49 So the first step is forced to state the null and alternative hypothesis.

00:54 So h -not is based on the fact that mew is equals to 8 .6 and the alternative hypothesis.

00:58 Is that meal is greater than 8 .6.

01:02 So it's time for us to get our test statistics which can either be a z test or a t test.

01:07 I'm going to be using a z test because we have the value of the population standard division and also the sample size is actually greater than 100 obeying the central limit theorem.

01:16 So we have our z to be equals to x bar minus meal divided by zima divided by the script of n.

01:23 X by in this case of us is 7 .7 minus our mule is 8 .6 divided by the sigma is 2 .0 divided by the scrote of 100.

01:33 So when we will substitute all of this into the formula, we have 7 .7 minus 8 .6 divided by 2 .0 divided by the square root of 100 and that is a minus 4 .5...

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