Christina is the senior vice president for Nursing Services...

Christina is the senior vice president for Nursing Services at St. Michael General Hospital. Recently, she noticed in job posting for nurses that those who are unionized seem not to offer the same wages. She decided to investigate and gathered the following information. Groups Union Nonunion Mean Wage/hour $20.75 $18.40 Sample Standard deviation 2.62 2.33 Sample Size 17 18 Would it be reasonable to reject what Christina noticed? Use $\alpha = 0.05$ level of significance; assuming that the variances are equal and the samples are random and independent, and the populations are normally distributed. (Please show and label all steps) do not use the p-value approach

Transcript

00:01 So you have data for union and non -union workers, and you have the mean, the standard deviation, and the sample size, and that was that 20 .75, the 5 .25, and the sample size of 40, and this is the $19 .80, assumingly, per hour, and $1 .90, and $45.

00:24 Now, the first question is they think that, and i'm going to put the hypotheses down, that the mean for the union is equal to the mean for the non -union.

00:37 And then alternately, we would think that the union job has a higher pay level than that of the non -union job.

00:45 So question one asked, is this a one - or a two -tail test? this is a one -tail test.

00:50 It's a right -tail test since they're using greater than.

00:53 For question two, we want to find what the decision rule is, and you're using an alpha level, alpha level of 0 .02, meaning that you need to put all 0 .02 in this upper level.

01:09 Now, you said something that it was a z value.

01:13 Technically, it's a t value because the sample standard deviation is not known, but then the t value, the degrees of freedom that we have here, here has to combine these two together and there is an ominous formula that we usually is software to find or we use the lesser of the two sample sizes and just estimate it to be 39.

01:37 There's a way of pooling it that you add these two together, 85 and then take away two and get 83 if we use pooling.

01:45 And again, that ominous formula that i never have my students memorize will give you something probably somewhere close to this.

01:52 So you'll some, textbooks say if these numbers are both greater than equal to 40 to go ahead and use a z and it sounds like that's what you're being told to do is to have this as a z value so the z value that has 2 % in the upper tail you really have a table in the back of your book most likely that tells that but that that z value is about 2 .054 you can also use your inverse normal this area down here has to be 0 .98 and find it to decimal place like that, or you could estimate it in between.

02:28 So that's the critical value.

02:30 And so if z is larger than that value, then that would cause you to reject the null.

02:38 Now, your third question asks you to look at what the test statistic is, and you have it as a z value.

02:45 Again, technically it's a t value.

02:48 And we would take the difference between these two means, and then we divide that by the square root of, and we take that.

02:57 5 .25 squared over 40 plus that 1 .5 squared over 45...

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