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A survey at a school where you played a financial game to see how much you can make money shows: Male students receive a financial gain of 29.5% on average_ The sur...

Question

A survey at a school where you played a financial game to see how much you can make money shows: Male students receive a financial gain of 29.5% on average_ The survey included 22 female students and they received a financial gain of 35.1% on average and have a standard deviation of 4.8%_Assume that the financial gain is a normally distributed variable and investigate using a hypothesis test whether female students have a higher financial gain margin than male students_ Use two methods. You can

A survey at a school where you played a financial game to see how much you can make money shows: Male students receive a financial gain of 29.5% on average_ The survey included 22 female students and they received a financial gain of 35.1% on average and have a standard deviation of 4.8%_ Assume that the financial gain is a normally distributed variable and investigate using a hypothesis test whether female students have a higher financial gain margin than male students_ Use two methods. You can choose one method yourself. The second should be the p-value method. Use both methods to draw your conclusions. Show all your calculations clearly:



Answers

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section. In a survey of 3005 adults aged 57 through 85 years, it was found that $81.7 \%$ of them used at least one prescription medication (based on data from "Use of Prescription and Over-the-Counter Medications and Dietary Supplements Among Older Adults in the United States," by Qato et al., Journal of the American Medical Association, Vol. 300, No. 24 ). Use a 0.01 significance level to test the claim that more than $3 / 4$ of adults use at least one prescription medication. Does the rate of prescription use among adults appear to be high?

Were given that significance level. Alfa is equal to zero point. You're one. And people who have pulled in favor can be represented by X, which is equal to 481 on the people who opposed is 401. So we get the total number of people this 401 plus 481 physical to 100. You need to Okay, okay, on the given claim is equal to zero point five. Sure. Yeah. Okay, So the claim is that either the null hypothesis or alternative high boxes state the opposite each other on the null hypothesis needs to contain the value mentioned claim. Okay, so you know that the null hypothesis is P is equal to 05 Walter type boxes is just a P is not you. Does your there for the sample proportion, which is the number of successes divided by sample size is equal to X over and X number of successes on. And there's a number simple slips So to your 481 of 382 which is approximately equal to 0.54 54 now, if to determine value of the test statistics of Z is equal to p. There is pure not or the square root of P not times one minus over. And so therefore, if you're plugging all the values since Europe 5454 here minus peanut which is 050 divided by the square root off June 5 times one might in Super five, which is a stupid can over 882 which is equal to 2.7. It's you. So now the P value is the probability of obtaining the value of the test statistic or value more extreme. When the hypothesis is true, you have to to determine the value of P using the normal probability table. Okay, on the former P is equal to P or C was negative. 2.7. Yeah, where is he? Greater than 2.7, which is equal to two p. See Destiny 2.7, which will give us a final answer of two times your point shoes. You 35 which is Joe could choose your shit. Sure. Okay, so now, since this P value is less than our, um, significant level Alfa, we can reject the null hypothesis because PVA is less than significance lovely. Therefore there is sufficient evidence to reject the claim that the proportion of subjects responded in favor is equal to 0.5 and therefore the politicians claim is false.

So the objective is the test to claim that 51 0.2% of newborn babies or boys So let P be the proportion off newborn babies there, boys. Therefore, we confined our hypothesis, which is your 0.5120 on our alternative hypothesis, which is the P is not equal, right? Did you have been fine? Two's your and let the level of significance Alfa P equals 0.4 Okay, so now we know that the number of births in the sample is 160. The number of newborn babies there, boys is you got to for 26 2 different example. Proportion of newborn babies, daughter boys, this will be had is equal to X for her end, which is equal to 426 over 160 which is equal to 0.4 95 Okay, now the Z test statistic he Yeah, school of peak over and cute is equal to one Manus people on p European five tissue. So if you just go ahead and plug these values and P and bullshit, we will get as he the statistic of negatives. You 0.9 eight So now, from the alternative hypothesis, it is clear that the test is two tailed on the standard normal table. At 5% level of significance, you can find the two tailed critical value to B C off to 1.96 Therefore, that is a critical value off plus or minus six, and you're the value of the test statistic is negative 0.98 which is greater than the critical value of negative 1.96 So therefore we can fail to reject the hypothesis and conclude that the proportion of newborn babies are 51.2%.

All right for this question they are asking us to find first the null hypothesis in the alternative hypothesis is so we have a claim that the freshman students on a campus study at least 2.5 hours per day. The null hypothesis is that claim so freshman study at least greater than equal to 2.5 hours for a day. And the alternative hypothesis is what the statistics cost. They're skeptical of this claim. So the alternative hypothesis is that freshman studied less than 2.5 hours per day. So what we need to do is we need to figure out is this null hypothesis Is this claim true? In order to do that, we need to find the p value. So in order to find the P value, the first thing we need to do is find the test statistic on that shows us how many standard deviations away from the mean is the results. So our T statistic, we find that with the formula of the, um, sample me and minus the population mean over the sample standard deviation over the square root of the sample size. So when we put that information in, Um, we're gonna go ahead and convert the minutes in the problem. Two hours. Um, to do that, you just divide by 60. So we have our 137 minutes becomes too point to a hours minus 2.5 hours over the sample standard deviation, which is, um, 0.75 hours, 45 minutes. It's 450.75 hours divided by the square root of the sample size, which is 30. And when you put that in your calculator, you get negative 1.6 07 Now that we have our t score, um, we need to find the P value. To do that, all you need to do is look up a tea table. You can even just google that and you'll find a tea table. You find negative 1.6 of seven, and it will give you the P value. The P value that you are given with that, um, to score is 0.6 now. What this means is that the probability of getting 137 minutes as the means so that's the 2.28 hours in the study is basically a 6% chance if the null hypothesis is true. So in order to decide about this study, we need to know what Alfa is. Alfa is the significance level. So Alfa is the probability of rejecting the null hypothesis when in fact it's actually true. The Alfa that they gave us is 0.1 which means that 1% of the time on the threshold is 1% of the time. You would reject the normal hypothesis when, actually it's true. Our P value, as we've already found, is 0.6 If the P value is higher than the Alfa, then that means that we do not reject the NOL. We keep the null. So go ahead and write for our d um, that we failed to rejection and all the P value. When the pea value is high, then all flies. When the people you is low, the norm must go. So that's a little saying to help you remember. So since the P value is greater than Alfa, then re failed to reject. So our conclusion in summary is that the student claim that freshman study at least two enough hours today is correct hours per day, and what it means is that we at least it's true. Um, what it means is that at least we don't have enough evidence to say that that's not true. So we're going to say that, according to this study, that the null hypothesis we cannot reject it, that it is still true that freshman students study at least 2.5 hours pretty.


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