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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

Short Answer

Expert verified

Part (a): Both conclusions are same.

This means that the conclusion for confidence interval is same as the conclusion for hypothesis test.

Part (b): Both conclusions are same.

This means, that the conclusion for confidence interval is same as the conclusion for hypothesis test.

Step by step solution

01

Part (a) Step 1. Given information.

Consider the given question,

The null hypothesis isH0:μ=μ0.

The alternative hypothesis isHa:μ>μ0.

02

Part (a) Step 2. Compute the confidence interval.

On using the MINITAB procedure,

  1. Choose Stat>Basic Statistics>1-Sample t.
  2. In Samples in Column, enter the column of Distance.
  3. In Perform hypothesis test, enter the test mean as 180.
  4. Check Options, enter Confidence level as 95.
  5. Choose greater than in alternative.
  6. Click OK in all dialogue boxes.

Test of μ=180vs>180

Hence, from the MINITAB output, the 90% lower bound is 180.42.

Here, the population mean is 180, which lies below the 95% lower bound. Then, the null hypothesis is rejected at5%level.

03

Part (a) Step 3. Using the Hypothesis test.

Using the Hypothesis test

From Problem 9-144E, the P-value is 0.031. The P-value is lesser than the level of significance. Therefore, the null hypothesis is rejected.

It can be concluded that the club will hit the ball more than 180 years.

Hence, both conclusions are same. It means, that the conclusion for confidence interval is same as the conclusion for hypothesis test.

On computing the 99% confidence interval,

  1. Choose Stat>Basic Statistics>1-Sample t.
  2. In Samples in Column, enter the column of Distance.
  3. In Perform hypothesis test, enter the test mean as 180.
  4. Check Options, enter Confidence level as 99.
  5. Choose greater than in alternative.
  6. Click OK in all dialogue boxes.
04

Part (a) Step 4. Take one sample T.

Hence, from the MINITAB output, the 99% lower bound is 178.91.

Here, the population mean is 180, which lies above the 99% lower bound. Then, the null hypothesis is not rejected at 1% level.

Using the Hypothesis test,

From Problem 9-144E, the P-value is 0.031. The P-value is greater than the level of significance. Therefore, the null hypothesis is not rejected.

Thus, there is no evidence to infer that the club will hit the ball more than 180 years.

Hence, both conclusions are same. It means, that the conclusion for confidence interval is same as the conclusion for hypothesis test.

05

Part (b) Step 1. Compute the confidence interval.

On computing the 95% confidence interval,

  1. Choose Stat>Basic Statistics>1-Sample t.
  2. In Samples in Column, enter the column of Angle.
  3. In Perform hypothesis test, enter the test mean as 45.
  4. Check Options, enter Confidence level as 95.
  5. Choose greater than in alternative.
  6. Click OK in all dialogue boxes.

Hence, from the MINITAB output, the 95% lower bound is 45.12.

Here, the population mean is 45, which lies above the 95% lower bound. Then, the null hypothesis is rejected at 5% level.

Using the Hypothesis test,

From Problem 9-145E, the P-value is 0.045. The P-value is lesser than the level of significance. Therefore, the null hypothesis is rejected.

Thus, the dat provide sufficient evidence to conclude that on average the angle between the body and head of an alligator during a death roll is greater than45.

Hence, both conclusions are same. It means, that the conclusion for confidence interval is same as the conclusion for hypothesis test.

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Most popular questions from this chapter

Betting the Spreads. 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 uniforn betting across teams, Las Vegas oddsmakers assign a point spread to each game. The point spread is the oddsmakers" prediction for th number of points by which the favored team will win. If you bet of 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-Seed?" (Chance, Vol. 11(1), pp. 27-31). 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 of10.9 points. For a particular game, a point-spread error of 0 indicates that the point spread 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 answer in part (b).

The normal probability curve and stem-to-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.

In the given exercise, we have provided a sample mean, sample size, and population standard. In each case, use the one-mean z-test to perform the required hypothesis test at the 5% significance level.

x=20,n=24,σ=4,H0:μ=22,Ha:μ22

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 right- tailed test withα=0.01

We have been provided a sample mean, sample size, and population standard deviation. In the given case, use the one-mean z-test to perform the required hypothesis test at the 5% significance level.

x¯=23,n=15,σ=4,H0:μ=22,Ha:μ>22

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