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To improve the waiting time of the phone customer service by thebank, the manager randomly selects 49 clients and asks the numberof minutes for each of them waiting...

Question

To improve the waiting time of the phone customer service by thebank, the manager randomly selects 49 clients and asks the numberof minutes for each of them waiting for the customer service. Thedata are given below.7, 3, 2, 0, 2, 6, 13, 1, 18, 15, 2, 6, 34, 9, 5, 3, 6, 4, 2, 17, 9,4, 0, 4, 1, 13, 0, 2, 1, 40, 2, 8, 2, 1, 2, 5, 10, 9, 4, 2, 4, 5,6, 7, 0, 1, 17, 6, 4A)To construct a confidence interval for the mean number ofminutes for waiting based on the given number of sample size, whatassumpti

To improve the waiting time of the phone customer service by the bank, the manager randomly selects 49 clients and asks the number of minutes for each of them waiting for the customer service. The data are given below. 7, 3, 2, 0, 2, 6, 13, 1, 18, 15, 2, 6, 34, 9, 5, 3, 6, 4, 2, 17, 9, 4, 0, 4, 1, 13, 0, 2, 1, 40, 2, 8, 2, 1, 2, 5, 10, 9, 4, 2, 4, 5, 6, 7, 0, 1, 17, 6, 4 A)To construct a confidence interval for the mean number of minutes for waiting based on the given number of sample size, what assumption is underlying for the population distribution of the waiting time? 1 The number of minutes for waiting can be modeled by the bimodal distribution. 2 It does not matter what population distribution assumption is because the sample size is sufficiently large. 3 The number of minutes for waiting is normally distributed. 4The number of minutes for waiting has a right-skewed distribution. b)Find a 93% confidence interval for 𝜇μ, the mean number of minutes for waiting on the phone customer service. lower bound= upper bound= c)Calculate marrgin of error? d) What is the most appropriate interpretation of the confidence interval? 1We are 93% confident that the computed interval can capture the true parameter if we repeat the process of generating confidence interval from 49 sample observations. 2We are 93% confident that the mean of the waiting time minutes is within the interval from the answer in part (b). 3 There is a 93% chance that the client is unsatisfactory on their waiting time for the phone service. 4 93% of the waiting time minutes fall in the interval from the answer in part (b).



Answers

Suppose the following small data set represents a simple random sample from a population whose mean is 50 and standard deviation is $10 .$ $$\begin{array}{llllll}43 & 63 & 53 & 50 & 58 & 44 \\\hline 53 & 53 & 52 & 41 & 50 & 43\end{array}$$ (a) A normal probability plot indicates the data come from a population that is normally distributed with no outliers. Compute a $95 \%$ confidence interval for this data set, assuming $\sigma=10$ (b) Suppose the observation, $41,$ is inadvertently entered into the computer as $14 .$ Verify that this observation is an outlier. (c) Construct a $95 \%$ confidence interval on the data set with the outlier. What effect does the outlier have on the confidence interval? (d) Consider the following data set, which represents a simple random sample of size 36 from a population whose mean is 50 and standard deviation is $10 .$ $$\begin{array}{|llllll}43 & 63 & 53 & 50 & 58 & 44 \\\hline 53 & 53 & 52 & 41 & 50 & 43 \\ \hline 47 & 65 & 56 & 58 & 41 & 52 \\\hline 49 & 56 & 57 & 50 & 38 & 42 \\\hline 59 & 54 & 57 & 41 & 63 & 37 \\\hline 46 & 54 & 42 & 48 & 53 & 41\end{array}$$ Verify that the sample mean for the large data set is the same as the sample mean for the small data set. (e) Compute a $95 \%$ confidence interval for the large data set, assuming $\sigma=10 .$ Compare the results to part (a). What effect does increasing the sample size have on the confidence interval? (f) Suppose the last observation, $41,$ is inadvertently entered as $14 .$ Verify that this observation is an outlier. (g) Compute a $95 \%$ confidence interval for the large data set with the outlier, assuming $\sigma=10 .$ Compare the results to part (e). What effect does an outlier have on a confidence interval when the data set is large?

Question number 16 number on our hypothesis with state that the population mean new is equal toe the value mentioned in the claim. So it snowed two new equal 8.3 minutes. Alternative hypothesis State states the opposite off the null hypothesis according to the clean. Thus using less Zane, each one two new is losing 8.3 minutes. If the alternative hypotheses uses less, then and then the test is left field. If the alternative hypotheses is bigger, then then the test is right team. Okay, if the alternative hypothesis use is not equal, the test is toe steel question and will be the non hypertext with state. That's the population mean you is equal to the value mentioned in the claim, so it's no good. Two. New equal. 8.3 minutes. Alternative hypothesis states that the opposite off the null hypothesis according to the claims off using different or not equal So which one two new Not equal 8.3 minutes. Use alternative hypotheses uses less than so. Then the test is left. Didn't if the current of hypotheses uses figures in then the test is right feel if the alternative hypothesis is not equal, then the test is toe feeding. Question number seat. Final hypothesis would state that the population menu is equal to the value mentioned in the claim. So each mood two new equal, 4.5 minutes. Alternative hypothesis stated the opposite off the null hypothesis. According to the claim that's using more than so, each one to mule is bigger than 4.5 minutes if the alternative hypothesis images they lives in, so then the test is left. Field is the alternative. Hypotheses uses bigger than then This is right feeding. If the alternative hypothesis use is not equal, then the test is toe 18 200 meat. Final hypothesis with the steak that the population mean new is equal to the value mentioned in the claim. So each node, so new equal 4.5 minutes. Alternative hypothesis states is the opposite off the null hypothesis, according to the claim, thus using different or not equal, so each one tome you not equal 4.5 minutes in the alternative hypotheses uses less then then the test is left. Field is the current of hype because uses Biggers in, then just is right field. If the alternative hypothesis use is not equal, then this is toe field

What we want to conduct A p D. T. A pair differences tests at the alpha equals 1% confidence level. Testing the claim that the population means A bar next pr are not equal. We have the data for A and B. Given below assuming amount shapes, not your distribution on the right. I've already calculated D. Bar noted that an equal seven and calculated SD as 70.47 So we proceeded to do the five steps listed below to solve this first. We evaluate the requirements and hypotheses. So the requirements to use the students distribution have been met because the distribution shape we have degree of freedom and minus one equals six. Are null hypothesis is mute equals zero. Or alternative is beauty does not equal zero. And we're testing at alpha equals 00.1 confidence Next will compute the test statistic and P value. So the statistic is T equals D. Bar divided by SD over. Uh huh. 2.083 from a tea table. This puts p between .1.05. So we can conclude that P is greater than alpha, which means we fail to reject the null hypothesis, which means that we lack evidence that beauty does not equal to about.

So here we're looking at a study of average shopping times in a large national housewares store, which gives the following information, Um, about women with a female companion, women with a male companion and how long they shocked. And we're going to set up a statistical test to challenge the claim that women with a female friend spend an average of 8.3 minutes shopping in such a store. So for a what would we use for the null and alternative hypotheses if we believe that the average shopping time is less than 8.3 minutes? So the null hypothesis is just what we're testing? And an ultra hypothesis is anything that is an alternative to that So H known as the null hypothesis. And this would be meal is 8.3, and each one or ultra hypothesis would be new, does not equal 8.3. And then we're asked if this was a right tail left tailed or a two tailed test. And depending on the alternate hypothesis, operator greater than operator will be a right tail test. Less than operator is a left field test, and naughty cooperator is a two tailed test Yeah. Um, so we can see that. Yeah. Mhm. This would be a Okay, So our alternate hypothesis would be less than eight. Wait three minutes, So yeah. So we're proposing that the 8.3 minutes is incorrect, that we think it's less than that. So that means that this would be a left failed test. Okay. And then in bur asked, what would we use for the No, An alternate hypotheses if we believe the average shopping time is different from 8.3 minutes and what kind of test that is? So in this case, are no hypothesis would be the same 8.3, but are ultra hypothesis would just be new. It is not equal 8.3. And this is where we would get a two tailed test. Yeah, yeah, yes. And then continuing on stores that sell mainly, women should figure out a way to engage men. Um, suppose that an entertainment center was installed, and you now wish to challenge the claim that a woman with the male friend only spends 4.5 minutes shopping. So what would we use for our hypotheses? Um so are null Hypothesis would be u is equal to 4.5, and our alternate hypothesis would be that the average would be more than 4.5. And this would make that a right tail test and then for D were asked what we use for the null and alternative hypotheses. If we believe the average shopping time is just different, so are no hypothesis would be the same. 4.5 for alternate would be Neil does not equal 4.5. And that would make this a two tailed test, Yeah.

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.


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