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Chapter 9: Q. 18 (page 392)

In each part, we have given the value obtained for the test statistic, z, in a one-mean z-test. We have also specified whether the test is two tailed, left tailed, or right tailed. Determine the P-value in each case and decide whether, at the 5%significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.

a. z=-1.25; left-tailed test

b. z=2.36; right-tailed test

c. z=1.83; two-tailed test

Short Answer

Expert verified

For a. the data does not provide sufficient evidence, for b, the data provides sufficient evidence; and for c. the data does not provide sufficient evidence.

Step by step solution

01

Step 1. Given Information

The value obtained for the test statistic, z, in one-meanz-test.
The tests have been specified to be left-tailed, right-tailed and two-tailed respectively.

02

Step 2. Solving for a. 

The test static,z=-1.25{"x":[[4,32,32,5,5,32],[5,32],[51,78],[52,78],[99,119],[133,148,149,149],[161],[174,175,185,199,204,196,184,173,174,203],[241,213,213,212,213,222,234,240,241,240,235,216,211]],"y":[[51,51,51,115,116,116],[87,87],[73,73],[95,94],[83,83],[31,9,10,115],[115],[30,16,9,11,26,51,86,116,116,116],[9,9,9,51,51,48,51,63,88,107,116,116,97]],"t":[[0,0,0,0,0,0],[0,0],[0,0],[0,0],[0,0],[0,0,0,0],[0],[0,0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0,0,0,0,0]],"version":"2.0.0"} , which is left-tailed test.
P-value =P(Z鈮�z)

=P(Z鈮�-1.25)

=0.1056
The P-value is 0.1056, which is greater than5% level of significance.
That is, P-value>伪=0.05.
Thus, we fail to reject our null hypothesisH0.
Therefore, the data does not provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis or not at the 5%significance level.
03

Step 3. Solving for b.

The test static,z=2.36 , which is right-tailed test.
P-value =P(Z鈮�z)

=P(Z鈮�2.36)

=1-P(Z<2.36)

=1-0.9909

role="math" localid="1651239866313" =0.0091
The P-value is 0.0091, which is less than5% level of significance.
That is, P-value<伪=0.05.
Thus, we reject our null hypothesisH0.
Therefore, the data does provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis or not at the5%significance level.
04

Step 4. Solving for c.

The test static, z=1.83, which is two-tailed test.

P-value=2P(Z鈮�z)
role="math" localid="1651240615811" =2P(Z鈮�1.83)

=2[1-P(Z<1.83)]

=2[1-0.9664]

=2(0.0336)

=0.0672

The P-value is 0.0672, which is greater than 5%level of significance.That is, P-valuerole="math" localid="1651240116209" >伪=0.05.
Thus, we fail to reject our null hypothesisH0.
Therefore, the data does not provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis or not at the5% significance level.

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Most popular questions from this chapter

As we mentioned on page 378, the following relationship holds between hypothesis test and confidence intervals for one-mean z-procedures: For a two-tailed hypothesis test at the significance level 伪, the null hypothesis role="math" localid="1653038937481" H0:渭=渭0will be rejected in favor of the alternative hypothesis Ha:渭鈮�渭0if and only if 渭0lies outside the 1-伪-level confidence interval for 渭. In each case, illustrate the preceding relationship by obtaining the appropriate one-mean z-interval and comparing the result to the conclusion of the hypothesis test in the specified exercise.

Part (a): Exercise 9.84

Part (b): Exercise9.87

9.93 Cell Phones. The number of cell phone users has increased dramatically since 1987. According to the Semi-annual Wireless Survey, published by the Cellular Telecommunications & Internet Association, the mean local monthly bill for cell phone users in the United States was \(48.16in 2009 . Last year's local monthly bills, in dollars, for a random sample of 75 cell phone users are given on the WeissStats site. Use the technology of your choice to do the following.
a. Obtain a normal probability plot, boxplot, histogram, and stemand-leaf diagram of the data.
b. At the 5%significance level, do the data provide sufficient evidence to conclude that last year's mean local monthly bill for cell phone users decreased from the 2009 mean of \)48.16?Assume that the population standard deviation of last year's local monthly bills for cell phone users is $25.
c. Remove the two outliers from the data and repeat parts (a) and (b).
d. State your conclusions regarding the hypothesis test.

Dirt Bikes. Dirt bikes are simpler and lighter motorcysles that are designed for off-road events. Specifications for dirt bikes can be found through Motorcycle USA on their website www.motorcycle-usa.com. A random sample of 30 dirt bikes have a mean fuel capacity of 1.91gallons with a standard deviation of 0.74 gallons. At the 10%significance level, do the data provide sufficient evidence to conclude that the mean fuel tank capacity of all dirt bikes is less than 2 gallons?

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 with伪=0.05

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 蟽=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.
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