WEBVTT
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in question 51 were given that the average weekly pay
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for a woman with high school education was$520 and
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were asked to determine if the average is significantly higher
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for all women compared to women with only the high
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school diploma. Were also given a sample of 50
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working women in the file accompanying the textbook. And
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for part, they were asked to state or hypotheses
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to test our claim. And so we'll start with
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the alternative hypothesis. And that would be that the
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average weekly earnings that women is of all women in
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general is greater than$520 and therefore, the no
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hypothesis is that the mean is less than or equal
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to 520 for Part B. We're supposed to use
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the sample to find the sample average, the sample
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standard deviation, the test statistic and the P value
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. So I've looked at the file when I've calculated
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the sample average and that you know to 637 0.94
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a sample. Standard deviation was 148.47 and the next
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step is to calculate the test statistic. Since we're
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not given the populations standard deviation. Our sample average
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will follow the T distribution. So tea is there
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test statistic and recall that it is equal to the
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sample average minus the population average over the sample standard
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deviation divided by the square root of the sample size
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. And that comes out to you 5.62 So before
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we go to the tea table, we can note
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that our degrees of freedom is 49. We're gonna
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go to the tea table to look up a p
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value. So 49 degrees of freedom T score of
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5.62 is off the charts, which means that RPI
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value is actually smaller than 0.5 So we can scratch
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that and say p value is lesson 0.5 part C
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. We're told at the significance level of Alfa equals
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0.5 What is their conclusion regarding their test? And
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we can say that a P value, which is
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less than 0.5 is less than 0.5 which is Alva
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. So therefore, the P value is a lesson
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Alfa and therefore we reject the no hypothesis for Part
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D. We're told to repeat the hypothesis test using
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the critical value approach so the critical values were looking
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for the test statistic. Or that we're looking for
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the T score that corresponds to an area in the
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tail of 0.5 And so again, we go to
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our table for this. So an area that retail
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0.5 corresponds to 1.677 so we have critical value of
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one point 667 So that's 1.677 And we already have
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our test statistic from the previous part of the question
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that he was 5.62 And since that is greater than
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the critical value, we can reject the null hypothesis
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.