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We know the length of- lime between charges for the NiCad batteries for laptops follows normal distribution and the mean is 50 hours and the standard deviation is 1...

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We know the length of- lime between charges for the NiCad batteries for laptops follows normal distribution and the mean is 50 hours and the standard deviation is 15 hours. Ilavibmodnayakalxa owns computer and would like to know the probability that the length of time between charging will be between 45 and 55 hours What is the probability that the length of time between charging will be less than 40 hours? What is the probability that the length of time between charging will be more than 55 hou

We know the length of- lime between charges for the NiCad batteries for laptops follows normal distribution and the mean is 50 hours and the standard deviation is 15 hours. Ilavibmodnayakalxa owns computer and would like to know the probability that the length of time between charging will be between 45 and 55 hours What is the probability that the length of time between charging will be less than 40 hours? What is the probability that the length of time between charging will be more than 55 hours? 2) Suppose that you enter fishing contest The contest takes place in a pond where the fish lengths have norma distribution with mean 16 inches and standard deviation inches. Now suppose you want to know what length marks the bottom 10 percent of all the fish lengths in the pond What fish length are you looking for? The Daily $ & return follows normal distribution with mean of 0.00032 and standard deviation of 0.00859. What is the probability that Daily S & P return will be greater than 0.01? If we choose random sample of 20 days what is the sampling distribution? For these randomly chosen 20 days_ what is the probability that the rate of return will be larger than 0.005? Etaratuglgotek is researcher and she wants to determine the effectiveness of an exercise regimen. She gets 41 volunteers who participated in it and finds that for these volunteers the mean heart rate is 82.02 beats per minute and the standard deviation is 8.13 beats per minute_ Construct 90 % confidence interval for these volunteers_ Suppose we want to know the average age of a student at Bennistefjollneix College within 0 years_ We'd like to be 99 % confident about our result. From previous study we know the standard deviation to be 2.9 years_ Calculate the required sample size Ginilannarihcuritxanop has been told that The Dawrahdillabuzxma Company manufactures bulbs with an average time life which is 470 hours Ginilannarihcuritxanop thinks that the average life time is much less. He collects 100 light bulbs and finds that the mean is 465 hours deviation is 25 hours. Test with a standard Ginilannarihcuritxanop' significance claim at the " 5 % level of



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Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. (b) Find the value of the chi-square statistic for the sample. Are all the expected frequencies greater than $5 ?$ What sampling distribution will you use? What are the degrees of freedom? (c) Find or estimate the $P$ -value of the sample test statistic. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence? (e) Interpret your conclusion in the context of the application. Use the expected values $E$ to the hundredths place. Psychology: Myers-Briggs The following table shows the Myers-Briggs personality preferences for a random sample of 519 people in the listed professions (Atlas of Type Tables by Macdaid, McCaulley, and Kainz). T refers to thinking and $\mathrm{F}$ refers to feeling. $$ \begin{array}{l|c|c|c} \hline & \multicolumn{2}{c} {\text { Personality Preference Type }} & \\ \cline { 2 - 3 } \text { Occupation } & \multicolumn{1}{c} {\text { T }} & \text { Row Total } \\ \hline \text { Clergy (all denominations) } & 57 & 91 & 148 \\ \hline \text { M.D. } & 77 & 82 & 159 \\ \hline \text { Lawyer } & 118 & 94 & 212 \\ \hline \text { Column Total } & 252 & 267 & 519 \\ \hline \end{array} $$ Use the chi-square test to determine if the listed occupations and personality preferences are independent at the $0.01$ level of significance.

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.

We're told the population has mean new equals 8.8. And are given in the following sample data randomly selected from that population from that sample thing that we want to calculate the mean X bar and the sample standard deviation S. So to do so let's remember the formulas for each of these terms, sample mean X bar is defined as the sum of the data that I am in this case 7.36 And the sample standard deviation is defined as some of the deviation about the mean squared all divided by n minus one. In this case, 4.3 Next. What we want to do Is used the sample data to implement a two tailed test. That is we want to test whether or not the sample data suggests the actual meaning. The population differs from the nomen of 8.8 in either direction, higher or lower. We want to use significant level Alpha equals 0.5 And we note at this point that X. Is approximately normally distributed. So to answer this test question, we have to go through the following procedures in order to finalize The solution of this test. So first, what's the significance level? And hypotheses are physical. 0.05. No hypothesis H. Not musicals. 8.8. H. A mute does not equal the population. 8.8. What distribution are using completely associated test statistic? The distribution. We're going to use the student's T distribution because our population standard deviation sigma is unknown. We can definitely use this distribution though, because the shape is appropriate. Ak because X is approximately normally distributed, it's okay to use a student's T distribution. Next, what is the T stat stat is defined by this formula, which here we see, reduces down to T equals negative 1.337 Next let's use the T stat to compute the p and dribble and sketch this out on the student's T distribution. So since we have degree of freedom and minus one equals 13, we're going to use the two tailed T table, which you can find either on google and a staff textbook to identify what range of p values this T stat negative 1.337 falls between, we find it falls between 0.2 and 0.25 We can graph this as the area underneath the student's T distribution, which we highlighted here in yellow outside of our T stat. Uh In this case we have both negative 1.337 and positive 1.337 Because the detailed test next, what can we conclude from this? Well, he is greater than alba. So we have statistically insignificant findings, i. E. We cannot reject h not or no hypothesis. And that ultimately means that we lack evidence suggesting that the population mean differs from its no mean of a pointing

We want to conduct the pair differences. Test that alpha equals 5% confidence. Testing the claim that population mean expert A. Is greater than population index. Barbie. Were given pair data samples A and B. Here and we assume amount, shapes and distribution on the right. Have already concluded or rather computed the differences mundi bar sample size and and the standard deviation differences SD as 4.5, 10 and 4.12 respectively. With this info we can proceed to do five texas below to compute this test first. That the value of the requirements using student's T. Distribution as well as hypotheses. So the requirements on that because the distribution specified degree freedom and mine as well as nine are null and alternative hypotheses are nude equals zero moody greater than zero and we have alpha equals 00.5 is our confidence. Next. It's gonna be the test at the P value. The test that is T equals D. Bar over sdo route and equals 3.45 And the tea table. This gives P between 0.5 point 0005 Thus we can include since P is equal to alpha, that we reject the null hypothesis H not which means that we have evidence that moody is greater than zero.

Following is a solution to number 25 comparing to means for the amount of time lost due to hot temperatures Compared to the mean number need an amount of time lost due to disputes from superiors attitudes in the workforce. And the first part of this, all we're gonna do is just verify that the mean and the standard deviations for the two datasets are in fact 4.86 and 3.18 for the temperatures And then 65 and 288 for the attitudes. And it says use a calculator, so I'm using the T I T four and if you go to stat and then edit, you can see already put those numbers in. So L one represents, I think that's the temperature column and then L two is the attitude problem. And if you go back to stat and then cowpoke and then one bar stats, one variable stats the list, I'm just doing L one first and that is where we get that 4.86. And the standard deviation where this s is that's up 3.18. So that's where I get these numbers here, and then we'll do the same thing, cal one of our stats, but this time we're gonna do second to for L. Two, and we calculate that, and that's where we get the mean as 6.5, this X bar here, and then the standard deviation, we're not looking at signal, we're looking at essence is a sample sample standard deviation about 2.88 So that's where we get this 2.88 So we have verified that those are in fact the numbers, and now we're gonna do the two sample t test with the significance level of point oh five. Before we do that, we need to figure out what the um alternative hypothesis would be, because we already have the data. We can just punch that in after that and it's going to be a two tail not equal to test because it says one way or the other is just as are the two means different and whenever it doesn't specify which one is greater, you just assume that it's a two tail meaning not equal to so not equal to is going to be our alternative. Okay. And then there are no I didn't write it down but the Noles that they're equal to. Okay. So if we go to stat and then go over to tests and it's the two sample T test and since we already have the data, I'm just going to use the data instead of the summary stats because it's more accurate that way list one will be L one list too will be L to the frequencies can just stay as one and then the alternative the new one, let's change that to not equal to the pool is going to be no unless it's otherwise stated. And then we're going to calculate and this gives us everything. So the t. You know if you want you can put that in there. But really all I care about is this P value the P value is about 0.317 So I'm gonna write that down so the p value is equal to 2.317 And what we do is we explicitly compare the P value with our alpha value and this time the p value 0.317 is greater than our alpha value. Which means we fail to reject the null hypothesis. So any time the peabody is greater than alpha you fail to reject. If it's less than alpha than you actually reject. H not so if we fail to reject this null hypothesis, that means there's not enough evidence to say that these two means are different, or in other words, these two means appear to be the same, so I'm gonna write that out, or I'm gonna type it out because it's a little quicker. That way, I'm gonna, right, there is not sufficient evidence to suggest that the meantime lost due to hot temperatures is different from the meantime lost due to disputes from superiors attitudes in the workforce. Okay, so that is our two sample T test for these two population means.


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