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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:μ≠22

Short Answer

Expert verified

Ans: We have to reject H0at5%the level of significance as the value of z falls in the rejection region.

Step by step solution

01

Step 1. Given information.

given,

x—=20,n=24,σ=4,H0:μ=22,Ha:μ≠22

02

Step 2. Let's perform a hypothesis test.

H0:μ=22Ha:μ≠22

Where μis the population mean.

Since the alternative hypothesis contains the symbol localid="1651574793371" ≠.

And the test is a two-tailed test.

Now,

The level of significance is, α=0.05.

03

Step 3. Now, we find the hypothesis test about the mean, μ.

we have,

x—=20,n=24,σ=4

Test statistic

z=x¯−μ0σn=20−22424=−2.45

04

Step 4. Since this is a two-tailed test with α=0.05,  

By using the normal distribution table, the critical values are:

±zα/2=±z0.05/2=±z0.025=±1.96

Here the rejection regions are z<-zα/2orz>zα/2

i.e., z<-1.96orz>1.96

Here z=-2.45<-z0.025=-1.96

Since the level of significance as the value of z falls in the rejection region.

So, we have to reject localid="1651211012259" H0at5%.

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Most popular questions from this chapter

9.96 Left-Tailed Hypothesis Tests and Cls. In Exercise 8.105 on page 335, we introduced one-sided one-mean z-intervals. The following relationship holds between hypothesis tests and confidence intervals for one-mean z-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.
a. Exercise 9.85
b. Exercise 9.86

Serving Time. According to the Bureau of Crime Statistics and Research of Australia, as reported on Lawlink, the mean length of imprisonment for motor-vehicle-theft offenders in Australia is 16.7months. One hundred randomly selected motor-vehicle-theft offenders in Sydney. Australia, had a mean length of imprisonment of localid="1653225892125" 17.8months. At the localid="1653225896253" 5%significance level, do the data provide sufficient evidence to conclude that the mean length of imprisonment for motor-vehicle-theft offenders in Sydney differs from the national mean in Australia? Assume that the population standard deviation of the lengths of imprisonment for motor-vehicle-theft offenders in Sydney is localid="1653225901113" 6.0months.

Define theP- value of the hypothesis test.

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.

Left-Tailed Hypothesis Tests and CIs. 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=μ0 will be rejected in favor of the alternative hypothesis Ho:μ<μ0 if and only if μ0 is greater than or equal to the (1−α)-level upper confidence bounif for μ. In each case, illustrate the preceding relationship by obtaininy the appropriate upper confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.
a. Exercise9.117
b. Exercise9.118
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