For each of the following values of αfind the values of z for which H0:p1-p2=0would be rejected in favor of Ha:p1-p2<0.

aα=.01bα=.025cα=.05dα=.10

Short Answer

Expert verified

a. For α=.01,the value of Zis -2.326.

b. For α=.025,the value of Zis -1.96.

c. For α=.05 ,the value of Zis -1.645.

d. For α=.10,the value of Z is -1.28.

Step by step solution

01

Given information

We have to find critical values for the hypotheses

H0:p1-p2=0

And

Ha:p1-p2<0

02

Definition of Critical value

The critical value is the cut-off value for the test statistic, which decides whether to reject the null hypothesis or not.

03

Calculating Critical value

Here

α:The level of significance

α=.01

Since we have a left-tailed test (as an alternative hypothesis is left-tailed)

Using the standard normal table, the critical value at the 1% significance level for the left-tailed test is -2.326

That is

z=-2.326.

The null hypothesis will be rejected for any test statistic value less than -2.326.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Product failure behavior. An article in Hotwire (December 2002) discussed the length of time till the failure of a product produced at Hewlett Packard. At the end of the product’s lifetime, the time till failure is modeled using an exponential distribution with a mean of 500 thousand hours. In reliability jargon, this is known as the “wear-out” distribution for the product. During its normal (useful) life, assume the product’s time till failure is uniformly distributed over the range of 100 thousand to 1 million hours.

a. At the end of the product’s lifetime, find the probability that the product fails before 700 thousand hours.

b. During its normal (useful) life, find the probability that the product fails before 700 thousand hours.

c. Show that the probability of the product failing before 830 thousand hours is approximately the same for both the normal (useful) life distribution and the wear-out distribution.

Question: Consumers’ attitudes toward advertising. The two most common marketing tools used for product advertising are ads on television and ads in a print magazine. Consumers’ attitudes toward television and magazine advertising were investigated in the Journal of Advertising (Vol. 42, 2013). In one experiment, each in a sample of 159 college students were asked to rate both the television and the magazine marketing tool on a scale of 1 to 7 points according to whether the tool was a good example of advertising, a typical form of advertising, and a representative form of advertising. Summary statistics for these “typicality” scores are provided in the following table. One objective is to compare the mean ratings of TV and magazine advertisements.

a. The researchers analysed the data using a paired samples t-test. Explain why this is the most valid method of analysis. Give the null and alternative hypotheses for the test.

b. The researchers reported a paired t-value of 6.96 with an associated p-value of .001 and stated that the “mean difference between television and magazine advertising was statistically significant.” Explain what this means in the context of the hypothesis test.

c. To assess whether the result is “practically significant,” we require a confidence interval for the mean difference. Although this interval was not reported in the article, you can compute it using the information provided in the table. Find a 95% confidence interval for the mean difference and interpret the result. What is your opinion regarding whether the two means are “practically significant.”

Source: H. S. Jin and R. J. Lutz, “The Typicality and Accessibility of Consumer Attitudes Toward Television Advertising: Implications for the Measurement of Attitudes Toward Advertising in General,” Journal of Advertising, Vol. 42, No. 4, 2013 (from Table 1)

Question: Impact of race on football card values. Refer to the Electronic Journal of Sociology (2007) study of the Impact of race on the value of professional football players’ “rookie” cards, Exercise 12.72 (p. 756). Recall that the sample consisted of 148 rookie cards of NFL players who were inducted into the Football Hall of Fame (HOF). The researchers modelled the natural logarithm of card price (y) as a function of the following independent variables:

Race:x1=1ifblack,0ifwhiteCardavailability:x2=1ifhigh,0iflowCardvintage:x3=yearcardprintedFinalist:x4=naturallogarithmofnumberoftimesplayeronfinalHOFballotPosition-QB::x5=1ifquarterback,0ifnotPosition-RB:x7=1ifrunningback,0ifnotPosition-WR:x8=1ifwidereceiver,0ifnotPosition-TEx9=1iftightend,0ifnotPosition-DL:x10=1ifdefensivelineman,0ifnotPosition-LB:x11=1iflinebacker,0ifnotPosition-DB:x12=1ifdefensiveback,0ifnot

[Note: For position, offensive lineman is the base level.]

  1. The model E(y)=β0+β1x1+β2x2+β3x3+β4x4+β5x5+β6x6+β7x7+β8x8+β9x9+β10x10+β11x11+β12x12 was fit to the data with the following results:R2=0.705,Ra2=0.681,F=26.9.Interpret the results, practically. Make an inference about the overall adequacy of the model.
  2. Refer to part a. Statistics for the race variable were reported as follows:β^1=-0.147,sβ^1=-0.145,t=-1.014,p-value=0.312 .Use this information to make an inference about the impact of race on the value of professional football players’ rookie cards.
  3. Refer to part a. Statistics for the card vintage variable were reported as follows:β^3=-0.074,sβ^3=0.007,t=-10.92,p-value=.000.Use this information to make an inference about the impact of card vintage on the value of professional football players’ rookie cards.
  4. Write a first-order model for E(y) as a function of card vintage x3and position x5-x12that allows for the relationship between price and vintage to vary depending on position.

Shared leadership in airplane crews. Refer to the Human Factors (March 2014) study of shared leadership by the cockpit and cabin crews of a commercial airplane, Exercise 8.14 (p. 466). Recall that each crew was rated as working either successfully or unsuccessfully as a team. Then, during a simulated flight, the number of leadership functions exhibited per minute was determined for each individual crew member. One objective was to compare the mean leadership scores for successful and unsuccessful teams. How many crew members would need to be sampled from successful and unsuccessful teams to estimate the difference in means to within .05 with 99% confidence? Assume you will sample twice as many members from successful teams as from unsuccessful teams. Also, assume that the variance of the leadership scores for both groups is approximately .04.

Question: Independent random samples selected from two normal populations produced the sample means and standard deviations shown below.

Sample 1

Sample 2

n1= 17x¯1= 5.4s1= 3.4

role="math" localid="1660287338175" n2= 12x¯2=7.9s2=4.8

a. Conduct the testH0:(μ1-μ2)>10against Ha:(μ1-μ2)10. Interpret the results.

b. Estimateμ1-μ2 using a 95% confidence interval

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free