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Question 19 (5 points) survey of 781 U.S.adults with children conducted in 2009 by the Financial Industry Regulatory Authority, 294 said that they had saved money f...

Question

Question 19 (5 points) survey of 781 U.S.adults with children conducted in 2009 by the Financial Industry Regulatory Authority, 294 said that they had saved money for their children'$ collcgce education Can you conclude that more than 33% of U.S. adults with children have saved money for college? Use the 0.05 Ievel of significance. Find the P-Value and state conclusion:There cnough evidence t0 conclude that morc than 33% have savcd money for collegeThere is not enough evidence t0 conclude t

Question 19 (5 points) survey of 781 U.S.adults with children conducted in 2009 by the Financial Industry Regulatory Authority, 294 said that they had saved money for their children'$ collcgce education Can you conclude that more than 33% of U.S. adults with children have saved money for college? Use the 0.05 Ievel of significance. Find the P-Value and state conclusion: There cnough evidence t0 conclude that morc than 33% have savcd money for college There is not enough evidence t0 conclude that more than 33%6 havc savcd money for college: Pvalue 0.0058 P-value 2.7601 P-value



Answers

(a) identify the claim and state $H_{0}$, and $H_{a},(b)$ find the critical value(s) and identify the rejection region(s), (c) find the standardized test statistic $z,(d)$ decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Assume the samples are random and independent, and the populations are normally distributed. If convenient, use technology. A sociologist claims that children ages $6-17$ spent more time watching television in 1981 than children ages $6-17$ do today. A study was conducted in 1981 to find the time that children ages $6-17$ spent watching television on weekdays. The results (in hours per weekday) are shown below. Assume the population standard deviation is 0.6 hour. $$\begin{array}{llllllllll} 2.0 & 2.5 & 2.1 & 2.3 & 2.1 & 1.6 & 2.6 & 2.1 & 2.1 & 2.4 \\ 2.1 & 2.1 & 1.5 & 1.7 & 2.1 & 2.3 & 2.5 & 3.3 & 2.2 & 2.9 \\ 1.5 & 1.9 & 2.4 & 2.2 & 1.2 & 3.0 & 1.0 & 2.1 & 1.9 & 2.2 \end{array}$$ Recently, a similar study was conducted. The results are shown below. Assume the population standard deviation is 0.5 hour. $$\begin{array}{llllllllll} 2.9 & 1.8 & 0.9 & 1.6 & 2.0 & 1.7 & 2.5 & 1.1 & 1.6 & 2.0 \\ 1.4 & 1.7 & 1.7 & 1.9 & 1.6 & 1.7 & 1.2 & 2.0 & 2.6 & 1.6 \\ 1.5 & 2.5 & 1.6 & 2.1 & 1.7 & 1.8 & 1.1 & 1.4 & 1.2 & 2.3 \end{array}$$ At $\alpha=0.05,$ can you support the sociologist's claim?

Okay, So these are the statistics that are given to us. Now, what is going to be are null hypothesis. H nod. Now we are checking for the claim that the mean off the course evaluations is equal to four. So I will say that the mean of the population is equal to four, right? This is my age, not what will be my edge Alternative. My alternative will become new is not equal to four. Okay, now what is our sample size? 93. So what is going to be my degrees of freedom? My degrees of freedom is given by the formula and minus one which becomes 92. Okay, so how do we calculate the T statistic now? The formula that is required to calculate the T statistic is T is equal. Do expire minus mu mu. Is the hypothesized mean that we're checking for upon we do not know the population standard deviation. So we're going to use the sample standard deviation upon root end. Okay, So what is X bar now? What I'm going to do is I'm simply going toe put all of these values in the calculator, so I'm using a calculator. This is 3.91 minus four minus four divided by what is s 0.530 point 53 And then I have multiplied this with route 90. See, what is the sample size? It's 93. So multiplied by Route 93. Okay, so our test statistic turns out Toby minus 1.63 This is our P value minus 1.63 This is R P value. Okay, Now we have our value, and we also have our degrees of freedom, which is 92 because 93 minus one. We already found the degrees of freedom. Here it is. 92. This is all the information that we need to find the P values. What I'm doing is I'm using an online tool to find it. What was our T score? Minus 1.63 This is minus 1.63 What is our degrees of freedom? It is 92 we are checking for not equal to. So this is going to be a two tailed test. We hit calculate and the P value that we're getting a 0.1065 So r p value r P value R P vally is 0.1 650.1065 Okay, now this p value is greater than the Alfa. Okay, it is greater than 0.5 also and it is also greater than 0.1. So no matter what Alfa you take this p value is going to be greater than Alfa. And hence we will fail to reject Arnel hypothesis. We will fail to reject we fail to reject h not okay going up What was our h not R h not Waas Mu is equal to four. So we say that we do not have sufficient evidence to say that the mean is anything other than four and thes will be my answers.

Right. We have a data sample with n equals 38 r equals 21 corresponding to 21 correct options out of 38 corresponding to a certain problem we want to solve we want to test the claim that P for the population proportion is less than 0.67 with a significant level of 5%. This corresponds to a confidence level alpha equals 0.5 In order to conduct this hypothesis test will go through the following procedural steps following the identification of the confidence level just completed first. Is it appropriate to use the normal distribution to solve this? Yes. Both M P and Q P A greater than five. What are the hypotheses hypothesis is that H not P equals 50.67 H a p is less than 0.67 Ak We're using a one tailed test compute P hat and the test statistic P hat is all over. End 10.55 plugging that into RZ stat formula on the right yields negative 1.57 for Z. Next we use the computer P value. We use a Z table from which we find the P value is 0.582 That is the area under the normal curve left of our Z stat, as highlighted here on the right. Finally, we reject H not No, we do not because the p value is greater than alpha, and we interpret that to mean that we lack evidence that P is less than .67.

The following is a solution in over 14 goodness of fit test where we look at the distribution of customer ages in a snoop report and see if it gives it agrees with the report from the sample. So the observed data values here uh were given to you in the book. So 88 1 35 50 to 40 76 1 28. I went ahead and just copy those down. And then for the expected, you gotta take the percent Times the sample size to get the actual expected value. So 12% times 519, for example, is 62.28. So I did that for all six categories there for the different age ranges. So now we're ready to go ahead and To answer these questions. So what's the significance level that they give? They give you that in the problem? It's a 1% significance level. So alpha's .01. And then we also need to write down what are null and alternative hypotheses are now typically for goodness of fit, test the Noles that these two distributions are the same. And then the alternative is that they're not the same. So in this particular case, bringing it back to this example, we're going to say the distribution of customer ages, uh and the snoop report, I'm not sure what a stupid port is, agrees with the sample report. Right? And then the alternative is that they don't agree. So, the distribution of customer ages and the stoop report does not agree with those of the sample report. So now we need to find the chi square value, but before we do that, we need to make sure that the expected values are greater than five and they are. So the expected values already found them, and you just take the percent times the sample size and they're all greater than five. So that's one of those conditions for inference they need to be at least five and they are so we can use the chi square distribution because of that with five degrees of freedom. Now why is it that five degrees of freedom? Well because there are six categories 1234566 categories minus one is five. That's always gonna give you the degrees of freedom. So now we can find basically everything else with software. So I'm gonna use the T. IT. for like using the T. 84 for smaller data sets. Um But you can use Excel or are or SAs or mini tab whichever you prefer. Or you can certainly use the formula. It might take you a little longer but you can use the formula. So I went ahead and typed these in already but I put an L1 I put the observed data values so you can see him there and then an L. Two. I put the expected data values you can see him there. So what I like to do sometimes is look at the difference here and these look to be kind of far apart. At least some of them do. So I don't know. I think it's been close but I don't think this is going to be a very good fit. Well let's see. So if we're gonna stat and then tests and it's one of the last ones I went up first. It's the D option if you're using A T. I. T. For. And it's the chi square geo F. Test that stands for goodness of fit test. L one is your observed, assuming you put it in. L one. L two is the expected degrees of freedom. Was five. Remember? And then we calculate and that's gonna give us the two values that we need to. The chi square value. Is this first number right here? 15.651 15 651 And then the P. Value, I don't know if you saw it there but we'll go in round. It's about .008 0.8 So just what I thought it's less than alpha. So anytime the P values less than alpha we this is the beauty of the P value. Matthew explicitly compared the P value with the alpha value. And if it's less than alpha then you always reject the null hypothesis. So we're going to reject h. Not so we're rejecting the claim that the distribution of customer ages and the snoop report agrees with the sample report. So that's how we conclude that will go and say there is. Oops. Well, I didn't mean to do that. Yeah, I can just type it in I guess. Mm. Sorry. Okay. I'll just I'll just actually write this out so there is okay, sufficient evidence to suggest the distribution of customer ages does not agree with that of the sample report. Okay. So these two distributions are not the same.

Okay, so we know that the null hypothesis h not is that p is equal to 0.75 on alternative hypothesis is that p is greater than the European 75 Okay, So to find the test statistic, we're gonna find pushing p. No, but says screw off peak, you ends. My cute is one minus p andi way. We know also from the question on 3000 five on DSO can calculate our Z value to be 8.48 Okay? By substituting all the values we are given. Okay. All right. So now the P value is equal to 0.0 She's lessons. European Ship one Therefore we can reject. Don't know. Hypothesis sentence. This is our critical.


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