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Why are mergers attractive to CEOs? One of the reasons might be a potential increase in market share that can come with the pooling of company markets. Suppose a random survey of CEOs is taken, and they are asked to respond on a scale from 1 to 5 (5 representing strongly agree) whether increase in market share is a good reason for considering a merger of their company with another. Suppose also that the data are as given here and that CEOs have been categorized by size of company and years they have been with their company. Use a two-way ANOVA to determine whether there are any significant differences in the responses to this question. Let  0.05.

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pH in Rain An environmentalist wanted to determine if the mean acidity of rain differed among Alaska, Florida, and Texas. He randomly selected six rain dates at each of the three locations and obtained the following data:

(a) State the null and alternative hypotheses.

(b) Verify that the requirements to use the one-way ANOVA procedure are satisfied. Normal probability plots indicate that the sample data come from a normal population.

(c) Test the hypothesis that the mean pHs in the rainwater are the same at the  level of significance.

(d) Draw boxplots of the pH in rain for the three states to support the results obtained in part (c).

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Putting It Together: Women, Aspirin, and Heart Attacks In a famous study by the Physicians Health Study Group from Harvard University from the late 1980s, 22,000 healthy male physicians were randomly divided into two groups; half the physicians took aspirin every other day, and the others were given a placebo. Of the physicians in the aspirin group, 104 heart attacks occurred; of the physicians in the placebo group, 189 heart attacks occurred. The results were statistically significant, which led to the advice that males should take an aspirin every other day in the interest of reducing the chance of having a heart attack. Does the same advice apply to women?
In a randomized, placebo-controlled study, 39,876 healthy women 45 years of age or older were randomly divided into two groups. The women in group 1 received 100 mg of aspirin every other day; the women in group 2 received a placebo every other day. The women were monitored for 10 years to determine if they experienced a cardiovascular event (such as heart attack or stroke). Of the 19,934 in the aspirin group, 477 experienced a heart attack. Of the 19,942 women in the placebo group, 522 experienced a heart attack.

(a) What is the population being studied? What is the sample?

(b) What is the response variable? Is it qualitative or quantitative?

(c) What are the treatments? (d) What type of experimental design is this?

(e) How does randomization deal with the explanatory variables that were not controlled in the study?

(f) Determine whether the proportion of cardiovascular events in each treatment group is different using a two-sample Z-test for comparing two proportions. Use the  level of significance. What is the test statistic?

(g) Determine whether the proportion of cardiovascular events in each treatment group is different using a chi-square test for homogeneity of proportions. Use the  level of significance. What is the test statistic?

(h) Square the test statistic from part (f) and compare it to the test statistic from part (g). What do you conclude?

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Untensity of Ponda forogramare pe week or to come an Anonduce posing and loyal Gatorant of whom today buy the program follow tribution for each ind promises to a recycling chat pays only 10 per program. The standard deviation 6.000 grams and meet regram 52 Rotero the stand somalt for a) What is the cost of understag demand wat tog Casad your response.place for University football programs are printed 1 week prior to each home game. Attendance averages 60,000 screaming and loyal fans, of whom two-thirds usually buy the program following a normal distribution, for $4 mal each. Unsold programs are sent to a recycling center that pays only 20 cents per program. The standard deviation is 5,000 programs, and the cost to print each program is $1. a) What is the cost of underestimating demand for each program? C = Cost of shortage (underestimated) = Sales price/unit – Cost/unit = $4 – $1 = S 3 b) What is the overage cost per program? Co = Cost of overage (overestimated) = Cost/unit – Salvage value/unit = $1 – $ 20 = $ 0.80 c) How many programs should be ordered per game? Determine the mean for the number of sold programs using that only two-thirds of fans usually buy the program HE – ().60.0 (60,000) = 40.000 (Enter your response as a whole number.) Now compute the service level, that is, the probability of not stocking out Service level cs 3 C + C 3 +0.80 0.7895 (Enter your response as a real number rounded to four decimal places.) decimal cs 3 Service level = 0.7895 (Enter your response as a real number rounded to four decimal C+C 3 + 0.80 places.) Therefore, find the Z score for the normal distribution that yields a probability of 0.7895. Hence, 78,95% of the area under the normal curve must be to the left of the optimal stocking level. a H = 40,000 Stocking 6 = 5,000 level 78.95% For an area of 0.7895, the Z value is 0.805 (enter your response as a real number rounded to three decima/ places.) Determine the correct formula to find the optimal stocking level. Choose the correct answer below. A. Optimal stocking level = (H + 0).Z B. Optimal stocking level = (u – 0).Z c. Optimal stocking level = + 2.0 D. Optimal stocking level = H- Given u = 40,000, 6=5,000, and Z = 0.805, compute the optimal stocking level. Optimal stocking level = 40,000 + (0.805)(5,000) = 44,025 (Enter your response as a whole number.) d) What is the stockout risk for this order size? Recall that the service level is the complement of the probability of a stockout. Stockout risk = 1 – Service level = 1 -0.7895 = 0.2105 (Enter your response as a real number rounded to four decimal places.) Therfore, 44,025 programs should be printed to each home game. 1f 44.025 programs are printed, the stockout risk is 0.2105 or 21.05%.

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