Following questions are answered (See attached word file for full list of 34 Questions)::
Question 1. [12 marks]
Market research has indicated that customers are likely to bypass Roma tomatoes that weigh less than 70 grams. A produce company produces Roma tomatoes that average 74.0 grams with a standard deviation of 3.2 grams.
(a) [2 marks] Assuming that the normal distribution is a reasonable model for the
weights of these tomatoes, what proportion of Roma tomatoes are currently undersize (less than 70g)?
(b) [2 marks] How much must a Roma tomato weigh to be among the heaviest 10%?
(c ) [2 marks] The aim of the current research is to reduce the proportion of
undersized tomatoes to no more than 2%. One way of reducing this proportion is to reduce the standard deviation. If the average size of the tomatoes remains 74.0 grams, what must the target standard deviation be to achieve the 2% goal?
(d) [3 marks] The company claims that the goal of 2% undersized tomatoes is
reached.To test this, a random sample of 25 tomatoes is taken. What is the
distribution of undersized tomatoes in this sample if the company’s claim is true?
Explain your reasoning.
In an article in Marketing Science, Silk and Berndt investigate the output of advertising agencies. They describe ad agency output by finding the shares of dollar billing volume coming from various media categories such as network television, spot television, newspaper, radio, and so forth.
Suppose that a random sample of 400 U.S. advertising agencies gives an average percentage share of billing volume from network television equal to 7.46 percent with a standard deviation of 1.42 percent. Further, suppose that a random sample of 400 U.S. advertising agencies gives an average percentage share of billing volume from spot television commercials equal to 12.44 percent with a standard deviation of 1.55 percent.
Using the sample information, does it appear that the mean percentage share of billing volume from spot television commercials for the U.S. advertising agencies is greater than the mean percentage share of billing volume from network television? Explain.
Module #3: Sampling Distributions, Estimates, and Hypothesis Testing
 Identify which of these types of sampling is used: random, systematic, convenience, stratified, or cluster.
a) The instructor of this course observed at a Walnut Creek Police sobriety checkpoint at which every fifth driver was stopped and interviewed. Some drivers were arrested.
b) The instructor of this course observed professional wine tasters working at a winery in Napa Valley, CA. Assume that a taste test involved three different wines randomly selected from each of five different wineries.
c) The U.S. Department of Corrections collects data about returning prisoners by randomly selecting five federal prisons and surveying all of the prisoners in each of the prisons.
d) In a Gallup poll, 1003 adults were called after their telephone numbers were randomly generated by a computer, and 20% of them said that they get news on the Internet every day.
e) The instructor of this course surveyed all of my students to obtain sample data consisting of the number of credit cards students possess in one of my statistics classes.
 In March 16, 1998, issue of Fortune magazine, the results of a survey of 2,221 MBA students from across the United States conducted by the Stockholm-based academic consulting firm Universum showed that only 20 percent of MBA students expect to stay at their first job five years or more. Source: Shalley Branch, “MBAs: What Do They Really Want,” Fortune (March 16, 1998), p.167.
a) Assuming that a random sample was selected, construct a 98% confidence interval for the proportion of all U.S. MBA students who expect to stay at their first job five years or more.
b) Based on the interval from a), can you conclude that there is strong evidence that less than one-fourth of all U.S. MBA students expect to stay? Explain why.
 An earlier study claims that U.S. adults spend an average of 114 minutes with their families per day. A recently taken sample of 25 adults showed that they spend an average of 109 minutes per day with their families. The sample standard deviation is 11 minutes. Assume that the time spent by adults with their families has an approximate normal distribution. We wish to test whether the mean time spent currently by all adults with their families is less than 114 minutes a day.
a) Construct a 95% confidence interval for the mean time spent by all adults with their families.
b) Does the sample information support that the mean time spent currently by all adults with their families is less than 114 minutes a day? Explain your conclusion in words.
 When 40 people used the Weight Watchers diet for one year, their mean weight loss was 3.00 lb. (based on data from “Comparison of the Atkins, Ornish, Weight Watchers, and Zone Diets for Weight Loss and Heart Disease Reduction,” by Dansinger, et al., Journal of the American Medical Association, Vol. 293, No. 1). Assume that the standard deviation of all such weight changes is s = 4.9 lb. We shall use a 0.01 significance level to test the claim that the mean weight loss is greater than 0.
a) Set up the null and alternative hypotheses, and perform the hypothesis test.
b) Based on these results, does the diet appear to be effective? Does the diet appear to have practical significance?
Given: µo = 70 grams, µ=74 grams, sigma=3.2 grams
P(X< 70 grams) = P (z < (74-70)/3.2) = P( z < 1.25) = normsdist(1.25) = 0.894 Ans.
We know that 90% of the area under the normal curve from the start point is:
Normsdist ((X-74)/3.2) = 0.90
We also know that critical value of z which divides the normal curve in the proportion of 0.9 and 0.1 is: NORMSINV ((0.5-0.9/2)*2) = 1.2816
Hence equating both the equation, we have
(X-74)/3.2 = 1.2816 OR x = 1.2816*3.2 + 74 = 78.1011 grams Ans.
Roma tomato should weigh at least 78.1012 grams to be among the heaviest 10%.
(c) Given: margin of error E = 0.02 or 2%, µ = 70.0 grams, sigma =?
Here we need to divide the normal curve in the proportion of 2% and 98% respectively.
The value of z which divides the normal curve in the proportion of 0.02 and 0.98 is: NORMSINV ((0.5-0.02/2)*2) = -2.0537
Now value of sigma can be found by following equation:
(70-74)/sigma = -2.0537 or sigma = 4/2.0537 = 1.9477 grams Ans.
To check: area beyond the z value of -2.0537= 1- normsdist (-2.0537) = 0.98 or 98%.
Also margin of error E = z*sigma = 2.0537*1.9477 = 4 grams and 74 – 4 = 70 grams is the undersized tomatoes. Hence area representing undersized potatoes is : normsdist(-2.0537) = 2%.
Given: n = 25, from (c) we know relevant sigma = 1.9477 and z-value = -2.0537
Hence required distribution of this sample will be:
Mean = 74 grams (point estimate) Ans.
Sample standard deviation s = sigma/sqrt (n) = 1.9477/sqrt (25) = 0.3895 grams Ans.
Margin of error E = z*s = 2.0537*0.3895 = 0.7999 grams
Hence 74 – 0.7999 = 73.2001 should be a point representing corresponding 2% area from the start of the normal curve, if company’s claim needs to be true. Or (73.2001-74)/0.3895 = -2.05366 which is correct as normsdist (-2.05366) =2%.
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