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For Exercises 2 through $12,$ perform each of these steps. Assume that all variables are normally or approximately normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Toy Assembly Test An educational researcher devised a wooden toy assembly project to test learning in 6-year-olds. The time in seconds to assemble the project was noted, and the toy was disassembled out of the child's sight. Then the child was given the task to repeat. The researcher would conclude that learning occurred if the mean of the second assembly times was less than the mean of the first assembly times. At $\alpha=0.01,$ can it be concluded that learning took place? Use the $P$ -value method, and find the $99 \%$ confidence interval of the difference in means. $$ \begin{array}{l|ccccccc}{\text { Child }} & {1} & {2} & {3} & {4} & {5} & {6} & {7} \\ \hline \text { Trial } 1 & {100} & {150} & {150} & {110} & {130} & {120} & {118} \\ \hline \text { Trial 2} & {90} & {130} & {150} & {90} & {105} & {110} & {120}\end{array} $$

Okay, so first thing we're looking at is what is the variable under consideration? And the answer to that would be height. Uh The two things that are being compared is cross fertilization and self fertilization. For part C. We are looking at the difference of heights and it's just gonna be, you know, a difference in heights. And then for D. Were asked why or why not have the numbers in the table paired differences? Uh Yes they are. That's what it says in the problem description. And then for e we're going to perform a 5% significance level test of inequality or just New one does not equal Mewtwo. Ah they tell us that D. is equal to 20.93. and SD is 37.74. So SD is 37-74 And the bar is 29 3. So we'll go ahead and put that there. Yeah. Mhm. Mhm. And whereas SD there, it is 37 74. What was our sample size I believe is 15, Yep. So I'll do the square root of 15. This is going to get us, let me tell you this in my calculator first. So we got Parentheses 20.93 Divided by 37 74 Divided by the Square Root of 15. So I get .143. Let me just make sure that still works. Yeah. Just put some parentheses right here, make sure that doesn't change the number. Yeah. Oh yeah it does. I forgot to put some parentheses what we actually have is 2.147. And I think the book will probably want us to simplify that to 215. Mhm. So then we can look here in table two for the 5% significance level We find AP value of .0497. And at the 5% significance level we can reject the hypothesis. However, at the 1% significance level, We will fail to reject the null hypothesis, and again, that's at alpha equals .01 and that's all our questions.

All right. So we're examining bolts and we have a sample size of 20. We want to conduct a chi square test at the 5% significance level and the sample standard deviation is 2.8875 mm. So here we are, I got a four data and next thing we can do this calculator degrees of freedom, which is going to be 20 minus one. So 19. Yeah, So we can look in table seven for 19 degrees of freedom at 5% uh significance because we're taking a right tail test, since the alternative hypothesis says greater than so, this critical value is going to be found in table seven back there. Yeah, we find it point oh five at 19 degrees of freedom 30.143 Mhm. Mhm. Okay. And then we can compare that to our evaluated value here the past where statistic and minus one is 19 as squared is going to be 28875 squared divided by fada squared so two squared. And this is going to get us a result of plugging the calculator here 19 Institute 0.8875 squared, divided by four, and we get 39.6. This result lands us to the right of this critical region, so we are in the rejection region. Therefore, we reject the null hypothesis. In favor of the alternative, there is sufficient data to suggest that the 20 bolts exceeds two seconds.

Okay. So what we're gonna be looking at is the comparison of I. Q. S. Of kids that went through some sort of lung cancer therapy and those that didn't. So what we're gonna be doing is taking a right tail test since our first group is bigger than the other one, the average is bigger for the other one. So we're gonna split this down right here and this is gonna be at the 1% significance level. So if our tests the state falls in this region we will reject the hypothesis. Mhm. So our test statistic is gonna be given by T. Equals this equation right here and we're going to substitute our values right there. So T. Is going to be equal to 84.4 minus 78.2. And then we're gonna take the sp which is the pool standard deviations sample size for the first one in 74 minus 1, 73. With a standard deviation of 12.6 squared. And for the other one it's 72 minus one. So 71 times the standard deviation of 15 squared. Yeah. And then we're gonna divide by the two adjusted sample sizes, Okay? Which is gonna be our degrees of freedom. Mhm. At 1% significant level. So we'll get a critical value later. But first our full statistic is 13.84 So go and put that in there. Take the square root of the solid reciprocal sample sizes. Mhm. And what we get for our t statistic is 2.71 And when we look in the back of the book we look between 102 100 degrees of freedom since 144 is about between them and we find out that it's 2.35 so obviously 2.71 is greater than 2.35 meaning that we do land in that critical region zone. So at least 1% significance level. We reject the null hypothesis. There is the this means that the accused of the kids that went through that therapy are lower.

For this exercise. We are given the following information. So we have these hypotheses and we're told that the true parameter P is p prime and P prime is less than the null hypothesis proportion. So therefore the alternative hypothesis is actually true now. For part, they were asked to show the following. Where's Ed is the test statistic for a one proportion test. So the one proportion test statistic is given as follows. Now, if we calculate the expectation on Zed, everything inside the brackets is constant except for the sample proportion. So the hypothesized proportion and and are both constant. So the expected value can be rewritten like this. This is because the expected value for a constant is that constant. Now, we also know from Chapter two that the expected value for a sample proportion is the true value. So this becomes and this is what we're trying to show, at least at least with respect to the expected value. So now, for the variants now, in our test statistic again, everything is a constant except for this sample proportion. So that is the only varying parameter. Now I will rewrite this like this, so I'm taking this constant here outside of the variance brackets. So therefore I squared. So that's why these square root symbol has disappeared. And we also know from Chapter two that the variance for a sample proportion is given by the following. So therefore we have the following result and this is what else we were trying to show in part A. So that completes part A. Now in part B, we want to show at the power of the lower tailed test is that is we want to find the probability of getting a test statistic lesson or equal to negative of the critical value when the alternative hypothesis is actually true. Now, if our test statistic is that and we have just found its expected value and it's variants, we would expect that this value is normally distributed as standard normally distributed because we are taking a parameter, we're subtracting its mean from it and then normalizing by its standard deviation. So to find the power of the test, we're looking for the probability that the standard normal distributed parameter is less than negative, the critical value and now just plugging in our values for the standard deviation, which is the square root of the variance of the test statistic and the expected value for the test statistic. And if you look at what's asked for in question be when you were asked to show this right here now for Part C were given this information. So we have some hypotheses and were given the true proportion as 0.8 and we're testing at a significance level of 0.1 This means that are critical. Value is negative, 2.3 to 6, and we are also given n equals 225. And so we know that we know that the alternative hypothesis is actually true and were asked, What is the probability that our test will detect this? What is the probability that our test will result in us rejecting the null hypothesis? In other words, what is the probability of us getting a test statistic less than minus 2.3 to 6? So all we have to do here is plug in all the numbers into this formula, and this gives us a value of 0.978 So the power of this test, the probability of rejecting the null hypothesis when the alternative hypothesis is true, is 0.978


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