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Name-_____________________________________________________In "Ages of Oscar-Winning Best Actors and Actresses" (Mathematics Teacher Magazine by Richard Brown and Gretchen Davis), the authors compare the ages of actors and actresses at the time they won Oscars for best actor or actress. The results for recent winners from each category are listed in the following table. Output for both men’s and women’s ages are provided. Men’s AgesTI 83/84 Outputx=45.1Σx=1625Σx2=77141sx=10.4σx=10.3n=36minX=31Q1=37.5Med=43Q3=52maxX=76Women’s AgesTI 83/84 Output83/84 Outputx=38.9Σx=1402Σx2=61022sx=13.6σx=13.4n=36minX=21Q1=30.5Med=35Q3=41.5maxX=80313843532131354232394455243135443239455626333749324045562633375033404660263438603541476027343961364248612834416137424862303441743743517630354180a. Sketch a box plot of men’s and women’s ages using the grid below.b. Compare the shapes of the two distributions, and identify any outliers and/or unusual values.The box plot for men’s age indicate one outlier above the other values which is numerically different distant from other values. This outlier is 76 and tends to pull the sample mean upwards. The plot shows that most of the values are above the sample mean.The box plot for women’s age has three outliers above the population, i.e. 61, 74, and 80 and tends to pull the mean upwards. The plot shows that the values are almost evenly placed above and below the mean with more of them above the mean.c. Use appropriate measures of center and spread to compare the important features of the distributions of men’s and women’s ages. Please use complete sentences. Use back of page if necessary.With the data for men having an outlier, the mean may not be an appropriate measure of central tendency of the data. This is because it is pulled upwards by only one figure. However, median can be a relevant measure to compare the two populations. The median age for men is 43 while that of women is 35 and this indicates that women win Oscar awards at an early age than men. The measures of spread such as range also depicts the differences in the two populations. The range in men is 45 while that of women is 59. This suggests that women are more widely distributed in terms of the ages that receive Oscar awards. They start as early as 21 years and women as old as 80 years also get the award.2. An investigator obtains scores for job satisfaction and blood pressure from each of 10 employees. Higher scores for job satisfaction reflect higher levels of job satisfaction. 244602012954000 Job SystolicSatisfaction BP--------------- ---------34 12423 12819 15743 13356 11647 12532 14716 16755 11025 1562446020762000a. Mean and Standard Deviation for both variables.x= Σx /nFor job satisfaction,Σx= 350, n = 10x= 350 /10 = 35x= 35For systolic BP,Σx= 1363, n = 10x= 1363 /10 = 35x= 136.3Standard deviation = (xi-x )/nFor job satisfaction, the standard deviation is 13.86For systolic BP, the standard deviation is 18.26b. Calculate and interpret the correlation coefficientBy plotting the linear regression of the two variables, it emerges that R2 = 0.719Correlation coefficient = R2= 0.719= ± 0.848Based on the direction of the line, it would be negative, i.e. – 0.848By being negative, the correlation coefficient indicates that an increase in job satisfaction corresponds to a decrease in systolic blood pressure.c. Show a scatter plot with the regression line.d. If a person had a job satisfaction score of 43, what would you predict for the person’s systolic blood pressure and how does it compare with the observed value?Based on the equation of the regression line, y = -1.1177x + 175.42When x = 43, y would be calculated as follows:y = -1.1177 (43) + 175.42= 127.4Therefore, if a person has a job satisfaction score of 43, their systolic BP would be 127.4e. How would you interpret the slope and intercept of the regression line?The regression line has a negative slope, i.e. -1.1177 and this indicates that an increase in one variable causes a decrease in the other variable. The intercept of the regression is 175.42 is where it intersects the axis. Besides, the low value of the slope indicates that the slope is barely steep and the average rate of change is significantly small.f. Based on this information what conclusions are you entitled draw about the relationship between the two variables? Calculate and interpret the coefficient of determination and percentage of the blood pressure can be explained by the model/person’s job satisfaction? How would you rate the model?It can be concluded that the two variables are inversely related. An increase in one of the variable leads to a decrease in the other variable. The coefficient of determination from the regression line is 0.719. Therefore, we 71.9 percent of the blood pressure can be explained by the person’s job satisfaction.3. How many children do U.S. women give birth to during their childbearing years? The distribution of number of children X for women past their childbearing years appears below.Number of Children X012345 or moreProbability P(X)0.1930.1740.344?0.074.034a. What is the probability that a randomly selected woman past childbearing age has 3 children?P (woman past childbearing age has 3 children) = 1 – (Sum of all the other probabilities)P (woman past childbearing age has 3 children) = 1 – [0.193 + 0.174 + 0.344 + 0.074 + 0.034)P (woman past childbearing age has 3 children) = 1 – 0.819P (woman past childbearing age has 3 children) = 0. 181b. Find the probability that a randomly selected woman has two or less children.P (woman has two or less children) = ...........?P (woman has two or less children) = [ P (X = 0) + P (X = 1) + P (X = 2)P (woman has two or less children) = [ 0.193+0.174 + 0.344]P (woman has two or more children) = 0.711c. Find the probability that a randomly selected woman has more than 1 child but less than 5 or more children.P (woman has more than 1 child but less than 5 or more children) = 1 - [P (X = 0) + P (X = 1) + P (X = 5 or more)]P (woman has more than 1 child but less than 5 or more children) = 1 – [0.193 + 0.174 + 0.034)= 0.5994. Juanita Johnson has just invested $175,000 to open a new drive-up food store. If successful, she can expect an annual income of $75,000. If unsuccessful, she will lose $95,000. (The remaining $80,000 can be recovered by selling the equipment.) If the probability of success is 0.85, find her mathematical expectation for the first year.The probability of success is 0.85 while that of failure is 0.15. If she succeeds, she gains an income of $75,000. If she loses, the loss will be -$95,000. An expression to establish her expectation in the first year will be as follows:75,000 (0.85) + -95,000 (0.15) = $49,5005. The fuel consumption of a Boeing747 in cruising mode averages 3213 gallons per hour. Assume that the consumption is normally distributed with a standard deviation of 180 gallons per hour. If a 747 is in cruising mode, what is the probability the fuel consumption is (a) between 3000 and 3500 gallons per hour?Mean =??= 3213Standard deviation =??= 180Let X be the random variable that represents the consumption is normally distributed.P( 3000 <X< 3500 )= P( ((3000-3213) / 180) <Z< ((3500-3213) / 180) )= P( -1.18 <Z< 1.59 )= P(Z<1.59) - P(Z<-1.18)= P(Z<1.59) - (1 - P(Z<1.18))= 0.9441 - ( 1 - 0.881)= 0.9441 - 0.119= 0.8251 (b) less than 3000 gallons per hour?P(X< 3000 ) = P(Z< ((3000-3213) / 180)= P(Z< -1.18 )= 1 - P(Z<1.18)= 1 - 0.881= 0.119 (c) more than 3500 gallons per hour?P(X> 3500 ) = P(Z> ((3500-3213) / 180)= P(Z> 1.59 )= 1 - P(Z<1.59)= 1 - 0.9441= 0.0559 (d) What does this information tell you about the fuel consumption of a 747?The fuel consumption of the airplane is mostly between 3000 and 35000 because it has the highest probability. (e) Would any of the above scenarios be considered unusual?There is no scenario that is unusual. An event is considered to be unusual if the probability of occurring is less than or equal to 0.05 (5%).6. A university wants to know more about the knowledge of students regarding international events. They are concerned that their students are uninformed in regards to news from other countries. A standardized test is used to assess students’ knowledge of world events (national reported mean=65). A sample of 30 students is tested (sample mean=58, Standard deviation of the mean for the sample=3.2). Compute a 99 percent confidence interval based on this sample's data. How do these students compare to the national sample?Confidence interval for Population mean is given as below:Confidence interval = x ± Z*σ/nFrom given data, we havex = 58σ = 3.2n = 30Confidence level = 99%Critical Z value = 2.5758 (by using z-table)Confidence interval = x ± Z*3.2/nConfidence interval = 58 ± 2.5758*3.2/30Confidence interval = 58 ± 1.5049Lower limit = 58 - 2.3514 = 56.4951Upper limit = 58 + 2.3514 =59.5049Confidence interval = (56.50, 59.50)From this confidence interval, it is observed that the students score is less than the national average.7. Researchers studying the effects of diet on growth would like to know if a vegetarian diet affects the height of a child. The researchers randomly selected 12 vegetarian children that are six years old. The average height of the children is 42.5 inches with a standard deviation of 3.8 inches. The average height for all six-year-old children is 45.75 inches. Conduct a hypothesis test to determine whether there is overwhelming evidence at α=.05 that six-year-old vegetarian children are not the same height as other six-year-old children. Assume that the heights of six-year-old vegetarian children are approximately normally distributed. (When answering this question, please state the null and alternative hypotheses, the claim, the test statistic value, the p value, confidence interval if appropriate, whether the Null Hypothesis is rejected and your response to the question.)The null hypothesis – Six-year-old vegetarian children ARE the same height as other six-year-old children.The alternative hypothesis – Six-year-old vegetarian children ARE NOT the same height as other six-year-old children.Ho: μ = 45.75Ha: μ ≠ 45.75α=.05 (as stated in the problem)Since we only selected 12 vegetarian children, we have a small sample size which means we use a t-test statistic.Degrees of freedom is n -1 = 12 – 1 = 11Since we are conducting a two-tailed test, our rejection region would have t-critical values of 2.201 and -2.201. These are gotten by looking at the t-table with 12 - 1=11 degrees of freedom and α=.025. The test statistic t is calculated as follows: t = 42. 5-45.753.812 = -2.963Since -2.963 < -2.201 we reject the null hypothesis in favor of the alternative hypothesisThere is significant evidence that, at the α=.05 level, vegetarian six year olds are not the same height as regular six year olds. ................
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