1. A mileage test is conducted for a new car model, the \u201cPizzazz.\u201d Thirty (n=30) random
\nselected Pizzazzes are driven for a month and the mileage is carefully measured in
\neach. The mean mileage for the sample is 28.6 miles per gallon (mpg) and the
\nsample standard deviation is 2.2 mpg. Estimate a 95% confidence interval for the
\nmean mpg in the entire population of Pizzazzes (you might need to round your
\nanswer a little bit to agree with mine).
\n2. Recent USA TODAY\/CNN\/Gallup Poll showed that most American people support
\nencryption on cell phones. The poll of 2000 adults was conducted and 1243 people said
\nthey supported technology companies to provide full encryption in cell phones.
\na) Find a 95% confidence interval for people said they supported technology
\ncompanies to provide full encryption in cell phones.
\nb) Perform the hypothesis test for 95 % confidence level that population supports the
\nfull encryption efforts from technology companies.
\n3. A random sample of 1,562 undergraduates enrolled in marketing courses was asked
\nto respond on a scale from one (strongly disagree) to seven (strongly agree) to the
\nproposition: “Advertising helps raise our standard of living.” The sample mean
\nresponse was 4.27 and the sample standard deviation was 1.32. Test the hypothesis
\nthat mean score is more than 4.
\n4. Of a sample of 361 owners of retail service and business firms that had gone into
\nbankruptcy, 105 reported having no professional assistance prior to opening the
\nbusiness. Test the null hypothesis with 99 % confidence that at most 25% of all
\nmembers of this population had no professional assistance before opening the
\nbusiness:
\n5. Pete\u2019s Powerful Pills (PPP) has branched out and is now promoting a vaccine which is
\nsupposed to prevent statisticitis, a nasty disease afflicting students at the local
\nEvermore University. They are trying to convince the university\u2019s health center that
\nthe vaccine is worth stocking. The campus health center has agreed to stock and
\nadminister the vaccine if it sells well enough, at least \u00b5 = 50 vaccines per week. PPP
\narranges for the health center to conduct a 36 week trial with the goal of convincing
\nthe center to stock the vaccine. They find that the center has administered an average
\nof X = 55 vaccines per week over that period with a standard deviation of s = 6.99.
\n(a) Compute a 95% confidence interval for the mean number of vaccines the health
\ncenter will administer per week in the long run. Does it look like the health center
\nwill be willing to stock the vaccine? Explain.
\n(b) Perform the hypothesis test using test statistics that PPP will conduct to convince
\nthe health center the vaccine is worth stocking. In other words, give the null and
\nalternative hypotheses, both mathematically and in words, and explain your
\nreasoning.
\n(c) Suppose that PPP makes a $5 profit on every vaccine administered by the health
\ncenter. Furthermore, suppose they need to average profits of at least $260 per week
\nfor it to be worthwhile to market the vaccine at the school. Can they be 95% sure of
\nmeeting their goal? Explain. (Hint: Use the CI computed in part (a)).
\n6. A statistics professor used X = \u201cnumber of class days attended\u201d (out of 30) as an
\nindependent variable to predict Y = \u201cscore received on final exam\u201d for a class of his
\nstudents. The resulting regression equation was Y = 39.4 + 1.4*X.
\nWhich of the following statements is true?
\na. If attendance increases by 1.4 days, the expected exam score will increase by 1
\npoint
\nb. If attendance increases by 1 day, the expected exam score will increase by 39.4
\npoints
\nc. If attendance increases by 1 day, the expected exam score will increase by 1.4
\npoints
\nd. If the student does not attend at all, the expected exam score is 1.4.
\n7. A large class of 360 students has just taken an exam. The exam consisted of 40 truefalse
\nquestions each of which was worth one point. A diligent teaching assistant has
\nrecorded the number of correct answers (Y) and the number of incorrect answers (X)
\nfor each student. Suppose that the student then regresses the variable Y on the
\nvariable X. What will be the values of model b0 (intercept), b1 (slope), R2 to fit to the
\ndata ?
\n8. A 95% confidence interval for b1 is determined to be (15,30). Interpret the meaning
\nof this interval.
\na) You can be 95% confident that the mean value of Y will fall between 15 and
\n30 units.
\nb) You can be 95% confident that the X value will increase by between 15 and
\n30 units for every one unit increase in Y.
\nc) You can be 95% confident that average value of Y will increase by between
\n15 and 30 units for every one unit increase in X.
\nd) At the 5% level of significance, there is no evidence of a linear relationship
\nbetween Y and X.
\n9. What do residuals represent?
\na. The difference between the actual Y values and the mean of Y.
\nb. The difference between the actual Yvalues and the predicted Y values.
\nc. The square root of the slope.
\nd. The predicted value of Y for the average X value.
\n10. A researcher wants to know if there is a relationship between the number of
\nshopping centers in a state and the retail sales (in billions $) of that state. A random
\nsample of 8 states is listed below. After determining, via a scatter-plot, that the data
\nfollowed a linear pattern, the regression line was found. Using the given data and the
\ngiven regression output answer the following questions.
\nState Num Sales
\n1 630 15.5
\n2 370 7.5
\n3 616 13.9
\n4 700 18.7
\n5 430 8.2
\n6 568 13.2
\n7 1200 23.0
\n8 2976 87.3
\na. What is the equation of the regression line?
\nb. Interpret the slope in the words of the problem.
\nc. Find r2 and interpret its meaning in the words of the problem.<\/p>\n","protected":false},"excerpt":{"rendered":"
1. A mileage test is conducted for a new car model, the \u201cPizzazz.\u201d Thirty (n=30) random selected Pizzazzes are driven for a month and the mileage is carefully measured in each. The mean mileage for the sample is 28.6 miles Read More …<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-237592","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts\/237592","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/comments?post=237592"}],"version-history":[{"count":0,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts\/237592\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/media?parent=237592"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/categories?post=237592"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/tags?post=237592"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}