Homework

Homework 5

Statistics Assignment Historically, it has been shown that the average amount of money a typical college spends during spring break is $160.00. The Daytona Beach Tourism Commission believes that the average amount of money spent increased during the most recent spring break. They surveyed 81 students and found that the mean daily spending over the most recent spring break was $166.00 with a sample standard deviation of $20.25. Presume that the money spent by college students during spring break is normally distributed. Is there sufficient evidence to suggest the average amount of money spent increased during the most recent spring break at the 10% level of significance?SolutionHo: µ= 160H1: µgt. 160 Critical z – value = 0.10Z = (166-160)/ (20.25/√81) = 2.666,From the z-table 2.666 gives a value of 0.996p-value= 1-0.996= 0.004Since 0.004lt. 0.10. Reject the null hypothesis: There is sufficient evidence to show that the spending increased during spring break.2) Historically, the average amount of time to assemble an electronic component on a production line has been 14 minutes. The supervisor of this production line is interested in determining whether this is no longer true. Assume that assembly time is normally distributed with a known population standard deviation of 3.4 minutes. The supervisor times the assembly of 25 randomly chosen components, and finds that the average time to assemble is 12.6 minutes. Is there sufficient evidence to suggest that the population mean assembly time is not 14 minutes at the 1% level of significance? What is the p-value?SolutionHo: µ= 14H1: µ≠14 Critical z – value = 0.010 Z = (12.6-14)/ (3.4/√25) = -2.059,From the z-table 2.059 gives a value of 0.0197p-value 0.0197*2= 0.0394 Since 0.0394gt. 0.010. Do not reject the null hypothesis: There is sufficient evidence to show that the population mean assembly is 14 minutes.3) The manufacturer of a new chewing gum asserts that at least 80% of dentists prefer their type of gum. An independent consumer research firm decides to test their claim. The findings of a sample of 200 dentists indicate that 76% of respondents actually prefer the manufacturers gum. At the 5% level, is there sufficient evidence to suggest that the population proportion of dentists who prefer the manufacturers gum is less than 80%? What is the p-value?Ho: p = 0.80H1: p lt. 0.80σ = √[ P * ( 1 – P ) / n ] = √[ 0.8 * ( 1 – 0.8) / 200 ] = 0.028z = (p – P) / σ = (.76 – .80)/0.028 = -1.43Looking for P(z=-1.43) in the z-table, we get 0.076P-value 0.076 Critical value = 0.05Since 0.076gt.0.05, do not reject the null hypothesis: There is enough evidence to suggest that the 80% of dentist prefer the chewing gum.4) A college professor is interested in determining the relationship between the number of hours a student sleeps prior to an exam and a students exam grade. The joint distribution is assumed bivariate normal. The professor draws a random sample of four students and records each students exam grade and hours of sleep preceding the exam. These data are found in the table below.Exam Grade , Hours of Sleep95 965 568 684 8The professor runs a simple regression, but has misplaced her regression output. Recognizing that Exam Grade is the dependent variable and Hours of Sleep is the sole independent variable, she asks you to perform the following:SolutionEquation of line y=mx+bM = b= XYXYX^2Y^2995855819025565325254225668408364624884672647056SUM28312226020624930a. What is the correlation between Exam Grade and Hours of Sleep?R2 = = 0.9721R=0.986b. What is the equation for the estimated regression line?M = = 7.6B= = -4.96Y= 7.6x-4.96c. What is the estimated exam grade when a student gets 7 hours of sleep?Y=7.6*7- 4.6= 48.6

Back To Top