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An airline wants select computer software package for its reservalion systcm Four software packages (1,2,3, and 4) are commercially available The airline will choos...

Question

An airline wants select computer software package for its reservalion systcm Four software packages (1,2,3, and 4) are commercially available The airline will choose the package that bumps as few passengers a9 possible during month An experiment is set uP in which each package used make reservations for randomly selected weeks: (A total of 20 weeks Was included in the experiment ) The number of passengers bumped each week is obtained_ which gives rise to the following Excel output:ANOVA Svurce %

An airline wants select computer software package for its reservalion systcm Four software packages (1,2,3, and 4) are commercially available The airline will choose the package that bumps as few passengers a9 possible during month An experiment is set uP in which each package used make reservations for randomly selected weeks: (A total of 20 weeks Was included in the experiment ) The number of passengers bumped each week is obtained_ which gives rise to the following Excel output: ANOVA Svurce %f Variation Between Groups Within Groups MS P-value 0.001474 F crit 3,238867 212.4 136.4 8304985 8,525 Total 348.8 10) Referring to Table 11-2, the among group (between-group) Iean squares is A) 70.8 B) 8.525 212.4 D) 637.2 11) Referring Table 11-2,at a significance level of 1%, A) there is sufficient evidence conclude that the mean number of customers bumped by the packages are all the same there is insufficient evidence to conclude that the mean number of customers bumped by the packages are notall the same: there is insufficient evidence conclude that the mean number of customers bumped by the - packages are all the same: D) there is sufficient evidence conclude that the mean number of customers bumped by the packages not all the same 12) professor would like to test the hypolhesis that the average number of minutes that sludent needs to complcte statistics exam is equal to 45 minutes Type error would occur if the professor concludes that the average eyan timc A) equal to 45 minutes when, in reality, the average time is less than 45 minutes_ B) equal to 45 minutes when, in reality, the average time is nol equal to 45 minutes not equal to 45 minules when, in reality, the average time is greater than 45 minutes D) not equal to 45 minutes when, in reality, the average time equal to 45 minules 13) Ten different families are tested for the number of gallons water day they use before and after viewing conservation vidco; Construct 90%a confidence interva" for the mean of the difference of the "before" minus the after"' times: Before 33 33 38 33 35 35 40 40 40 35 Aftcr 34 28 25 28 35 33 31 28 35 33 A) (3.47,5.35) B) (3.14,7. 88) C) (1.25, 6.35) D) (2.42, 7.98)



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.

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.

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

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.

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.


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