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Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty: To test if 68% also represents California percent fo...

Question

Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty: To test if 68% also represents California percent for full-time faculty teaching the online classes Long Beach City College (LBCC} in California was randomly selected for comparison In the same year; 34 of the 44 online courses LBCC offered were taught by full-time faculty: Conduct hypothesis test at the 5% level to determine if 68% represents California. NOTE: For more accurate result

Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty: To test if 68% also represents California percent for full-time faculty teaching the online classes Long Beach City College (LBCC} in California was randomly selected for comparison In the same year; 34 of the 44 online courses LBCC offered were taught by full-time faculty: Conduct hypothesis test at the 5% level to determine if 68% represents California. NOTE: For more accurate results use more California community colleges and this past year'$ data_ Note: If You are using Student's t-distribution for the problem_ You may assume that the underlying population is normally distributed, (In general; You must first prove that assumption, though:) Part (a} Part (b} Part (c) Part (d) State the distibution t use ior the test (Round your answers t0 four decimal places 0.68 0.32 Part (e) Part (f} What the p-value? (Round your answer to four decimal places_ 2675 Explain what the p-value means ior inis problem: If Hc is false, then there is chance equal to tne p-value that tne proportion of online courses taught by full-time faculty is not at least a5 different as the sample proportion from 6396 If Hc is true, then there iS chance equal to the p-value thai the proportion of online courses taugnt b} iull-time facuity not at least as different a5 ine sample proportion is irom 68%6_ # Ho true; then there chance equal to the p-value tha the proportion of online courses taugnt by iuli-time facuity is at least as different as the sample proportion is from 68%6. If Hc is false, then there is chance equal to ine p-value that ine proportion of online courses taught by full-time faculty is at least as different a5 ine sample proportion Tom 68%_ Part (g} Part (h} Part (i) Construct 9590 confidence interval for the true proportion Sketch the graph of the situation: Label the point estinate and ine lower ad upper bounds of the confidence interval. (Round your answers t0 four decimal olaces 95% CI 669



Answers

(i) What is the level of significance? State the null and alternate hypotheses. (ii) Check Requirements What sampling distribution will you use? What assumptions are you making? What is the value of the sample test statistic? Compute the corresponding $z$ or $t$ value as appropriate. (iii) Find (or estimate) the $P$ -value. Sketch the sampling distribution and show the area corresponding to the $P$ -value. (iv) Based on your answers in parts (i)-(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level a? (v) Interpret your conclusion in the context of the application.Note: For degrees of freedom $d . f$. not in the Student's $t$ table, use the closest $d . f$. that is smaller. In some situations, this choice of $d . f .$ may increase the $P$ -value a small amount, and therefore produce a slightly more "conservative" answer.Answers may vary due to rounding. Bell Peppers The pathogen Phytophthora capsici causes bell peppers to wilt and die. Because bell peppers are an important commercial crop, this organism has undergone a great deal of agricultural research. It is thought that too much water aids the spread of the pathogen. Two fields are under study. The first step in the research project is to compare the mean soil water content for the two fields (Source: Journal of Agricultural, Biological, and Environmental Statistics, Vol. 2, No. 2). Units are percentage of water by volume of soil.Field A samples, $x_{1}:$ $$\begin{array}{llllll}10.2 & 10.7 & 15.5 & 10.4 & 9.9 & 10.0 \\ 15.1 & 15.2 & 13.8 & 14.1 & 11.4 & 11.5\end{array}$$.Field B samples, $x_{2}:$$$\begin{array}{rrrrrrrr}8.1 & 8.5 & 8.4 & 7.3 & 8.0 & 7.1 & 13.9 & 12.2 \\13.4 & 11.3 & 12.6 & 12.6 & 12.7 & 12.4 & 11.3 & 12.5\end{array}$$.Use a calculator with mean and standard deviation keys to verify that $\bar{x}_{1} \approx 12.53, s_{1} \approx 2.39$ $\bar{x}_{2} \approx 10.77,$ and $s_{2} \approx 2.40$ (a) Assuming the distribution of soil water content in each field is mound-shaped and symmetrical, use a $5 \%$ level of significance to test the claim that field $A$ has, on average, a higher soil water content than field B. (b) Find a $90 \%$ confidence interval for $\mu_{1}-\mu_{2}$. Explain the meaning of the confidence interval in the context of the problem.

The following is solution video to number 26 which is two sample T tests comparing meantime loss in the workforce with stressors and intimidators. And you just got to be really careful here. The only kind of weird thing is They kind of switch up the order on you so X one represents the intimidators. In the next two represents the stressors. And we want to see if the stressors, so I'm actually gonna write that down. So this is intimidators and this is stressors. Okay, So it says is there enough evidence to suggest that the stressor is greater than the intimidator? So keeping that in mind, we're going to write less than some you won Is less than you two, Even though it says greater than but the orders just switched, if that makes sense? So the greater than in this case means less than for our alternative, since the order is opposite. Okay, so the first part of this is just verifying that these means. And standard deviations are four and 2.38 for intimidators, and then 5.5 and 2.784 stressors. So I went ahead and it says to use a calculator, so I'm using a T 84 if you go to stat and then edit here are the list, so L one is the intimidator and then L two is the stressor. So if you go back to stat over to couch and then it's one of our stats, we can change that to L one and this will be the four and the 238 that was over here. So four and 238 And then go back to stat couch. One bar stats, change that to L two calculate, and that's where we get to 55 and the 27855 to 78 So we verified that that is the way to do it. So then the second part of it is to kind of sort of not formally conduct a hypothesis test but essentially come up with the conclusion here and it is gonna be a two sample T test. We're gonna use the calculator since we already have the data input anyway. So it's two sample T test for the alpha is point oh five. And the alternative we already talked about, the alternative is less than not greater than but less than so. Um if you go back to stat and then we're gonna air over two tests and it's the fourth one down is the T. Test. Two sample T test because we don't know the standard deviation for the population, We don't know sigma, we only know as um so sincerity of the data, I'm just going to keep it as data instead of doing the summary stats. L one is the first list. L two is the second list and you can keep those frequencies as one and then we're gonna change this alternative to less than you two. And then pulled is usually no unless they tell you otherwise and then we're gonna go and calculate and you can have that test statistic there if you want but you really don't need to because the p value is all you need. So it's 0.14. So let's write that down. So the p values 0.14 and what we do is we explicitly compare the P value with the alpha. In this case 0.14 is greater than point oh five. So that means whenever the p values greater than alpha, we fail to reject H not so we failed to reject h. Not whenever the p values greater than alpha at the p values less than alpha, then we will reject the null hypothesis. So keep in mind that the null hypothesis is generally always saying that these two means are equal. We can we're failing to reject that, so we're accepting that to be the truth. So let's go ahead and complete this hypothesis test. And we're gonna say there is not sufficient evidence to suggest that the mean time lost due to stressors is greater than the meantime lost due to intimidators. That's spelled right? I have no idea. Looks like it's not so intimidators. Okay, that's why math teachers aren't spelling enthusiast. So intimidators um now you could also say there is not sufficient evidence to suggest that the meantime lost due to intimidators is less than the meantime loss due to stressors. If you want to keep to that less than thing. I just wrote it like this because that's the way the book had it. So there is not enough evidence to say that these two means are different or that one means more than the other. So we're accepting the truth that these two means are probably about the same.

Solution number 17. And this is a, an interesting problem with the Poisson distribution. And the Poisson distribution is the type of discrete distribution where the events are random and rare. And uh, this, in this case it's a traffic accidents, Daily traffic accidents, and there is an average, that's what we use this lambda for. So λ,, which is the average of the mean is 1.72 accidents per day. Were asked to find the probability that zero accidents occur and the probability that one occurs To occur three occurs and then greater than four. Curse. Now you can use the formula, but again, I like to use the uh, software. So what I'm gonna do is I'm gonna go to second distribution and then I'm gonna go down to the present pdf and then here it asks from you, or sometimes it lasts for lambda, and that's 1.72 And then the X value, I'm just going to find the probability that zero occurs, And that gives me .1791. I'm going to go in round here some point 1791. So you can do this with any type of software or you can just use the formula. Although the formula can take awhile .1791. And I'm gonna do it one more time. Just show you whether they're the second bars for distribution. And then I went to the Plaza pdf, right? That's the probability density function. So pdf. And the mu the mean is that's the land of 1.72. And this time we're gonna find the distribution or the probability that one occurs and it's about Point Let's say .308 point 308. And that's what you're gonna do for, you know, basically all the rest of them. Until you get to the greater than so zero point 2649, I'll go ahead and give you these answers here and then 0.15 one night. So then uh to get something that's greater than what you're gonna do is you're gonna take one go all the way up to infinity and save some time. We're just going to take one minus the four that we've already found. 0123 So there's another function in the calculator we can use is called the plaza CDF. The cumulative density function. And we're going to go up to three. So the CDF calculates the probability of zero plus the probability one plus probability two plus probability of three whenever you do CDF of three, so one minus that. Or you could just do one minus these four numbers here whichever you like. So one minus. And then 2nd distribution. And I'm going to go to the present CDF. And the main remember was 172 and then the X value. Now I'm not gonna put 0123 I'm just gonna put the three And it automatically calculates 0 1, 2 and three combined. And whenever you do that that should be your answer. So .0962 0962. You might get 61 if you just use these numbers here but you get the same thing if you do one minus and these all added together. Next up we find the expected value and any time you find the expected value just take the sample size times the probabilities. So for zero, remember there were 90 days that we looked at. So 90 times the probability of zero was remember that .1791? That should give you 16.119. And then all of these are gonna be the same 90 times something. And those some things are the probabilities. So .308 that's going to give you 27.72. So that means we can expect about 27.72 days where there are there's one wreck And then 90 times the point 2649. That should give us 23 841 And then 90 times .1519. He was this 13671. And the 90 times 0.0 962 Gives you 8658. Okay, so those are the probabilities. Those are the I'm sorry those are the expected values. And then the part c we find we're gonna find our use the good as fit test. So they observed that was the chart that was given and the expected we actually just found. So we're just gonna copy those expected values down 27.72 23-841 13.671 and then eight 658. So those are the expected values where we take the probabilities times the sample sizes. Okay? So now we can go back to our calculator and do the goodness of fit test. So if you go to stat and then edit you can see that here in L. One. I put the observed values and then L. Two I put the expected values. And then if you go to stat tests it's the chi square goodness of fit test. And the observed is L. One the expected sell to or you can use the formula or any other software. And degrees of freedom was four. So the degrees of freedom, that's actually another answer. The degrees of freedom of four. Since there are five categories there of 0123 and then greater than equal to four. So we can calculate that And that gives us a chi square value of about 12 509 So chi square equals 12.509 And then it also asked for degrees of freedom. The degrees of freedom, like I said, was equal to four because it's five minus one. Okay, so the p value let's look back at the P value, it's about .01. Let's go at some .014 and that is greater than the alpha, barely, but it is greater than the alpha. So whenever the p value is greater than the alpha than we fail to reject to reject H not. Which means in this case, actually, in all these goodness of fit test cases, the null hypothesis is that the distribution does in fact follow whatever we're talking about in this case, we're failing to reject that. So this follows a person distribution so the traffic accidents, and per day in this particular area, it does in fact follow a Poisson distribution.

All right, we have a population with no means 67 we want to calculate for the following sample the sample mean and the sample standard deviation So to do so, let's remember that the sample mean and sample standard deviation are defined by these formulas X. Bar is the some of the data divided by M. In this case 61.8 S is some of the deviations about the mean square all divided by n minus one or 10.6. Now, we want to implement a two tailed test, that is, we want to test whether or not from the sample we can find statistically significant suggesting the population means differs from the no mean of 67 ida direction. We use an alpha value of 670.1 and we note that X is approximately normally distributed. To implement this test, we have to answer the following questions to start. What's the significance of hypotheses after is equal to 0.1? No hypothesis, H and r is that mu is equal to 67? And our alternative hypothesis is that new is not equal to 67. What distribution of do we use? Let's compute the associated test statistic. The distribution will use the student's T. Because a population standard deviation sigma is unknown. We can use this distribution safely because the shape is described as normal, which is both mound shaped and symmetric from this. We have to derive a T stat defined by this formula. It reduces our negative 2.19 And this problem, let's compute the p interval from this. T stat. This is the degree of freedom is and minus one equals 15. We have the p p interval corresponding 150.75 PM. That's natures of 0.1. We find this p interval by using the two tailed T table, which can be found on google or a textbook for this T stat. The graph looks like this on the right where we have our absolute value, both negative and positive, much for artist at negative 2.19 and 2.19 And the P value is the area under the distribution to the left and right of these values. What can we conclude from these findings? Well, we concluded that since P is greater than alpha, we have statistically insignificant findings. Therefore, we can not reject h not and then we can't reject H not. We interpret this finding to mean that we lack any evidence suggesting that the population mean differs from its known mean of 67.

Problem. 20. The null hypothesis is that B one minus B two is equal to zero, and the alternative hypothesis is is that the one minus B two is bigger than zero, where be one proportion is X one over N ones for 30/50 she's open six and B two, which is equal to extra over in two, which is 22/50 which is Oh, point or four. The pooled proportion, which is X one plus x 2 30 plus 22 over in one plus in two she is 100. She's equal toe 1000.52 toe. Find the value off until the statistics, which is equal to the P one minus B two over a square root off 4.5 to 1 minus 4.5 52 square root off one over n. One plus one over and to which is equal to 1.6 eso. The probability is equal to the probability that that or the P value is equal, that the probability that that is bigger than 1.6, which is equal to the probability that set is smaller than negative 1.6, which is equal to open toe five, 48 and the P value is bigger than open toe five. So we say it to reject the null hypothesis is so there is no sufficient evidence to support support the claim.


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