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The normal probability curve and stem-and-leaf diagram of the data are shown in figure; σis 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 each of Exercises 9.41-9.46 ,determine the critical values for a one-mean z-test. For each exercise, draw a graph that illustrates your answer

A left-tailed test withα=0.05

Refer to Exercise 9.19. Explain what each of the following would mean.

(a) Type I error.

(b) Type II error.

(c) Correct decision.

Now suppose that the results of carrying out the hypothesis test lead to the rejection of the null hypothesis. Classify that conclusion by error type or as a correct decision if in fact the mean length of imprisonment for motor-vehicle-theft offenders in Sydney.

(d) equals the national mean of 16.7 months.

(e) differs from the national mean of 16.7 months.

Beef Consumption. According to Food Consumption, Prices,\and Expenditures, published by the U.S. Department of Agriculture. the mean consumption of beef per person in 2011 was 57.5 lb. A sample of 40 people taken this year yielded the data, in pounds, on last year's beef consumption given on the Weiss Stats site. Use the technology of your choice to do the following.

a. Obtain a normal probability plot, a boxplot, a histogram, and a stem-and-leaf diagram of the data on beef consumptions.

b. Decide, at the 5% significance level, whether last year's mean beef consumption is less than the 2011 mean of 57.5 lb. Apply the one mean t-test.

c. The sample data contain four potential outliers: 0, 0, 0, and 13.Remove those four observations, repeat the hypothesis test in part (b), and compare your result with that obtained in part (b).

d. Assuming that the four potential outliers are not recording errors, comment on the advisability of removing them from the sample data before performing the hypothesis test.

e. What action would you take regarding this hypothesis test?

In Exercise 8.146 on page 345,we introduced one-sided one mean-t-intervals. The following relationship holds between hypothesis test and confidence intervals for one-mean t-procedures: For a right-tailed hypothesis test at the significance level α, the null hypothesis H0:μ=μ0will be rejected in favor of the alternative hypothesis data-custom-editor="chemistry" Ha:μ>μ0if and only if μ0is less than or equal to the 1-α-level lower confidence bound for μ. In each case, illustrate the preceding relationship by obtaining the appropriate lower confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.

Part (a): Exercise 9.114 (both parts)

Part (b) Exercise9.115

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.

Part (a): If, on average, the oddsmakers are estimating correctly, what is the (population) mean point-spread error?

Part (b): Use the data to decide, at the 5%significance level, whether the (population) mean point-spread error?

Part (c): Interpret your answer in part (b).

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