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The normal probability curve and stem-to-leaf diagram of the data are shown in figure; σis unknown.

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

Neither 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 small. Moreover, the distribution of the variable is not normally distributed. Hence, neither test is used for given scenario.

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Most popular questions from this chapter

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

Part (a): Exercise 9.117

Part (b): Exercise 9.118

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? They obtained the difference between the actual margin of victory and the point spread, called the point-spread error for \(2109\) college basketball games. The mean-point spread error was found to be \(-0.2\) point with a standard deviation of \(10.9\) points. For a particular game, a point spread error of \(0\) indicates that the point was a perfect estimate of the two teams relative abilities.

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

b. Use the data to decide, at the \(5%\) significance level, whether the (population) mean point-spread error differs from \(0\).

c. Interpret your answers in part (b).

How Far People Drive. In 2011, the average car in the United States was driven 13.5 thousand miles, as reported by the Federal Highway Administration in Highway Statistics. On the WeissStats site, we provide last year's distance driven, in thousands of miles, by each of 500 randomly selected cars. Use the technology of your choice to do the following.

a. Obtain a normal probability plot and histogram of the data.

b. Based on your results from part (a), can you reasonably apply the one-mean t-test to the data? Explain your reasoning.

c. At the 5 % significance level, do the data provide sufficient evidence to conclude that the mean distance driven last year differs from that in 2011?

Cheese Consumption. Refer to Problem 24. The following table provides last year's cheese consumption: in pounds, 35 randomly selected Americans.

4629333842403433323628472642363245243928334433263727313637373622443629

  1. At the 10%significance level, do the data provide sufficient evidence to conclude that last year's mean cheese consumption for all Americans has increased over the 2010 mean? Assume that σ=6.9lb. Use a z-test. (Note: The sum of the data is 1218lb.)
  2. Given the conclusion in part (a), if an error has been made, what type must it be? Explain your answer.

Refer to Exercise 9.17. 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 iron intake of all adult females under the age of 51 years.

(d) equals the RDA of 18 mg per day.

(e) is less than the RDA of 18 mg per day.

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