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The human resources department large organization wants t0 know which days of the week employees are absent in five-day work week Most would like t0 believe that em...

Question

The human resources department large organization wants t0 know which days of the week employees are absent in five-day work week Most would like t0 believe that employees are absent equally during the week: Suppose random sample of 125 managers in the organization were asked on which day of the week they had tne highes number of employee absences, and the results were distributed in the table below For the population of employees conduct goodness-of-fit test at 1% significant level to determine

The human resources department large organization wants t0 know which days of the week employees are absent in five-day work week Most would like t0 believe that employees are absent equally during the week: Suppose random sample of 125 managers in the organization were asked on which day of the week they had tne highes number of employee absences, and the results were distributed in the table below For the population of employees conduct goodness-of-fit test at 1% significant level to determine whether the days for tne highest number of absences occur with equal frequencies during five-day work week Monday Tuesday Wednesday Thursday Friday Number of Absences 35 What are the null and alternative hypotheses? Ho: The absent days occur with unequal frequences, that is, they do not fit a uniform distribution. Ha; The absent days occur with equal frequencies, that is, they do fit a uniform distribution. Ho: The absent days occur with equal frequences that they do fit a uniform distribution. Ha; The absent days occur with unequal frequencies_ that E they do not fit = unifomm distribution_ The absent days occur with equal frequences that IS, they do fit a uniform distribution Ha: The absent days occur wilh equal frequencies that tney do fit a uniform distribution Ho: The absent days occur wilh unequal frequences that is Ihey do not fit uniform distribution The absent days occur with unequal frequencies that they do not fit a uniform distribution: b) What are the expected (E) values of absenteeism for each day? Monday Tuesday Wednesday Thursday Friday Number of Absences



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. wanagement: Intimidators and Stressors This problem is based on information regarding productivity in leading Silicon Valley companies (see reference in Problem 21 ). In large corporations, an "intimidator" is an employee who tries to stop communication, sometimes sabotages others, and, above all, likes to listen to him-or herself talk. Let $x_{1}$ be a random variable representing productive hours per week lost by peer employees of an intimidator.$$\begin{array}{llllllll}x_{1}: & 8 & 3 & 6 & 2 & 2 & 5 & 2\end{array}$$. A "stressor" is an employee with a hot temper that leads to unproductive tantrums in corporate society. Let $x_{2}$ be a random variable representing productive hours per week lost by peer employees of a stressor.Use a calculator with mean and standard deviation keys to verify that $\bar{x}_{1}=4.00, s_{1} \approx 2.38$ $\bar{x}_{1}=5.5,$ and $s_{2} \approx 2.78$ (a) Assuming that the variables $x_{1}$ and $x_{2}$ are independent, do the data indicate that the population mean time lost due to stressors is greater than the population mean time lost due to intimidators? Use a $5 \%$ level of significance. (Assume that the population distributions of time lost due to intimidators and time lost due to stressors are each mound-shaped and symmetrical.) (b) Find a $90 \%$ confidence interval for $\mu_{1}-\mu_{2}$. Explain the meaning of the confidence interval in the context of the problem.

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.

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


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