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Chapter 9: Q. 31 (page 393)

The normal probability curve and stem-and-leaf diagram of the data are shown in figure; 蟽is known.

Perform Hypothesis test for mean of the population from which data is obtained and decide whether to use z-test, t-test or neither. Explain your answer.

Short Answer

Expert verified

z-test is used.

Step by step solution

01

Step 1. Given Information 

02

Step 2. Conditions to use z-test

Small Sample size:

If the sample size is less than 15, the z-test procedure is used when the variable is normally distributed or very close to being normally distributed.

Moderate Sample size:

If the sample size lies between 15 and 30, the z- test procedure is used when the variable far from being normally distributed or there is no outlier in the data.

Large Sample size:

If the sample size is greater than 30, the z- test procedure is used without any restriction.

03

Step 3. Conditions for t-test.

Small Sample size:

  • Samples are randomly selected from the population.
  • Population follows normal distribution or the sample size is larger.
  • The standard deviation is unknown.
04

Step 4. Explanation.

Here, the sample is selected from the population and the sample size is large. Moreover, the population standard deviation is known and there is no outlier. Thus, distribution of the variable is approximately normal. From the above conditions, it is clear that to use of the z- test procedure is appropriate.

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Most popular questions from this chapter

In the article "Business Employment Dynamics: New data on Gross Job Gains and Losses", J. Spletzer et al. examined gross job gains and losses as a percentage of the average of previous and current employment figures. A simple random sample of 20quarters provided the net percentage gains for jobs as presented on the WeissStats site. Use the technology of your choice to do the following.

Part (a): Decide whether, on average, the net percentage gain for jobs exceeds 0.2. Assume a population standard deviation of 0.42. Apply the one-mean z-test with a 5%significance level.

Part (b): Obtain a normal probability plot, boxplot, histogram and stem-and-leaf diagram of the data.

Part (c): Remove the outliers from the data and then repeat part (a).

Part (d): Comment on the advisability of using thez-test here.

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Use a t-test to decide, at the5%significance level, whether last year's mean value lost to purse snatching has decreased from the 2012mean. The mean and standard deviation of the data are 455.0and 86.8, respectively.

College basketball, and particularly the NCAA basketball tournament is a popular venue for gambling, from novices in office betting pools to high rollers. To encourage uniform betting across teams, Las Vegas oddsmakers assign a point spread to each game. The point spread is the oddsmakers' prediction for the number of points by which the favored team will win. If you bet on the favorite, you win the bet provided the favorite wins by more than the point spread; otherwise, you lose the bet. Is the point spread a good measure of the relative ability of the two teams? H. Stern and B. Mock addressed this question in the paper "College Basketball Upsets: Will a 16-Seed Ever Beat a 1-Speed? They obtained the difference between the actual margin of victory and the point spread, called the point-spread error, for 2109college basketball games. The mean point-spread error was found to be -0.2point with a standard deviation of 10.9points. For a particular game, a point-spread error of 0indicates that the point spread was a perfect estimate of the two teams' relative abilities.

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