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Average IS 327 . lest at 4 =.05JLI8 The average salary of a leading = Hollywood actor is $18 million. Test the accuracy of this claim with sample f 10 actors who ea...

Question

Average IS 327 . lest at 4 =.05JLI8 The average salary of a leading = Hollywood actor is $18 million. Test the accuracy of this claim with sample f 10 actors who earned an average of $21 million with standard deviation of $4 million Use & =.05.

average IS 327 . lest at 4 =.05 JLI8 The average salary of a leading = Hollywood actor is $18 million. Test the accuracy of this claim with sample f 10 actors who earned an average of $21 million with standard deviation of $4 million Use & =.05.



Answers

Salary An employer pays a mean salary for a 5-day workweek of $\$1250$ with a standard deviation of $\$129.$ On the weekends, his salary expenses have a mean of $450 with a standard deviation of $\$57.$ What is the mean and standard deviation of his total weekly salaries?

In this problem, we are looking at the random sample of starting salaries of 35 randomly selected um graduates of bachelor degrees and comparing it if the salary of last year is less than two years ago. So we would expect it to be equal to two years ago, which is 43,589. And we are testing if our new average is actually less than that. Um So again we calculate our Z score, Which are sample mean is 41202 Minour are expected mean of 43,589 divided by our standard deviation over our yes square root of our sample size. And then we'll compare that to our Z score that we've done and that gives us a Z score of negative 1.853. So it's definitely a good bit different when we look at it in comparison but then we have to determine at 5%. Is it different enough? Is it significantly different? So our Z score for 5% is negative 1.645. And in this case, Okay, so Here's our rejection region of 5%. Here's our 95. So where does our sample fall? Our sample falls in the rejection region At at negative 1.853. So in this case we would reject the knoll. It is significantly different enough to where we say we have evidence against the null hypothesis evidence that the salaries are actually smaller this year versus two years ago

In this question, we have a magazine publisher who says that the mean of the mean income of its readers Is $6,061,500. And the advertising agency believes that that is actually smaller. So one tell test here again, we have to calculate our Z. Score based off of our sample that the advertising agency has gathered And see if it is significantly less than are expected mean of 61,500. My guess looking at our standard deviation is that it's not going to be a standard deviation of 5850 and we're only, you know, 2000 ish off. Probably not significant. But we will finish the test here to see because we are dividing that standard deviation by the square root of 40. So it does get a good bit smaller there because we're looking at a sampling distribution, not just a regular distribution. And we get our Z score of negative 1.8379. It's right there on the on the border here And then our significance level is a 1% significance level. So we're gonna go to our appendix Point to one that is 2.3-6. So anything outside of uh in this case it's a one tailed test. Anything less than negative 2.3-6 would be um would be in the rejection region. So here's 1% of our data, here's 99 and our sample falls in the 99%. So we would fail to reject. R no R. C. Score is within range that it's not enough to reject it at that given significance level. Mhm.

Hello. Welcome to the video. And today we'll be talking about hypothesis tests and specifically how to know when our evidence does or does not sufficiently support our claim. So when this problem, we're told that the the average annual salary of nurses is $69,110. So we can translate that into no hypothesis by saying that mu or the average annual salary of nurses is equal to 69,110. And it is proposed that the true averages, um greater than greater than this value here. So we can translate that into an alternative hypothesis by saying that mu is greater than 69,110. So if we want Thio, visualize this here, let's see, we have ah distribution with a peak value at the proposed 69,110. Ah, with our with our hypothesis test here, we hope to find evidence to show that the true value of Mu is actually somewhere over here and not close to what is proposed here. So in other words, will be doing a right tailed hypothesis test you and we will be doing this hypothesis test at a 5% Calvillo. So, um, to test these hypotheses, we do a survey of 41 nurses. And so just you were told that a sample size of 41 um, we're told that the sample mean waas Ah, 71,000 121 with a sample standard deviation off. 7489. So this is the information we gathered from our sample. And with this information here, and the information we're given here, we can go to our calculators to find a testes, tick and P value to determine what we should do with our hypotheses. So if we go to our calculators, I'm using a J 84 plus and we go to the T test function, and we give it these parameters here. Um, so the proposed mean here is 69,110 the from our sample, the average was 71 1 21 sample standard deviation was 7489 with a sample size of 41. And just looking at our hypotheses, we hope to find a value that is greater. Then what is proposed here. So if we go to our calculators and give it all this information here and press enter were given a test statistic in this case a T score of 1.72 with the corresponding P value 0.46 0.47 So, uh, these are this is the information we're looking for here. And although it is really close, R P value is slightly less than our alfa level and therefore, our evidence is telling us that we should reject are no hypothesis and basically, in the context of the problem, our evidence is telling us that this value is not an accurate representation of mu. And instead I knew true value of Mu is actually somewhere greater than 69,110. It is actually probably somewhere up here rather than over here. But all in all, we reject the know have offices

In this question. We want to compare the average earnings of college graduates whether they're men or women, and is there a difference? And if there is a difference in income, what is that different in income between men and women? When he was hypothesis testing specifically the Z test presented at page 4 91 of the book because started Step one gather over data so we know that it started with the simple sites for one. So we know we simple the 100 men. So 100 men and 100 women and those simple we The average earning for men was 90 1002 $100. For women, it was lower. It was 57,000 $800. The standard deviation for all mental not just for a sample for the population of men is $15,000. For women, it's 12,000 $800. We're gonna right now, our hypotheses. So the no hypothesis is based on the question should be that the difference in price the difference in the present earnings or income is exactly $30,000. What is claimed in the question is that this difference is not $30,000 So we will have a two tailed tests as soon as I finish writing this. So the difference in income or earnings and not exactly sure the difference. I'm not economy major. So, uh, the difference is not $30,000 So we have a two tailed tests. Therefore, rejection area is this thing this area in red. So to tell our normal distribution now that we have our hypotheses, you have Oliver data our data that's going to step two. So step to want to find our critical value. We know that Alfa is 0.1 Therefore, Alfa over to is 0.0 Ah, another 05 So are critical values are plus minus yes. Close or minus two 0.575 minus 2.575 on the left of our graph and 2.575 on the right hand side of our graph ever critical value. Now we need a zeke to compare those critical values to. So Step three, figure out Z Z is the difference in earnings for our samples. Ah, I wrote it s because I was thinking simple, but it's w four women. Mine is the difference in earning for all men and all women divided by the variants in earnings for men divided by the number of men. Simple. Plus that the variants in earnings for women divided by the number of women. Simple. All of that inside a square root. Remember, H zero tells us that this quantity is difference for you men and women Home. Oh, minute on women is $30,000. So what we have for Z world be 90,000 200 minus 57,000 800 minus 30 1000 over 15,000 squared, divided by 100 plus 12,800 squared, Also divided by 100 Everything inside a square root. I will let you plug all those values in your calculator or your computer, and you will get a value of 1.21. Let's go compare debt Z value to the critical value found that step to so Z is 1.21 1 point. Anyone that say it's right here in any case is smaller than 2.575 So we're not in the rejection area, so we do not reject age zero. We accept with those samples we accept h zero. So since one since 1.21 is smaller, then 2.575 and we do not reject 80 which means the difference in mean income. Four man and women is 30 $1000 sell quick recap. We gathered all of her data for problem. We rolled down her hypotheses and figured out critical values to define a rejection area. Computed rz two see if it was in their rejection area and we decided, since it wasn't in the rejection area, we did not reject age zero and we concluded that the difference in mean and come for men and women is $30,000.


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