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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

Short Answer

Expert verified

The value of z is 0.97, critical value is 1.645,P=0.002and do not reject H0.

Step by step solution

01

Step 1. Given information.

Consider the given question,

x¯=23,n=15,σ=4,H0:μ=22,Ha:μ>22

02

Step 2. Consider the test hypothesis.

Consider the given hypothesis,

σ is the population mean.

The test hypothesis,

H0:μ=22VsHa:μ>22

Therefore, the test is right tailed test.

And the level of significance isα=0.05.

03

Step 3. Use the test statistics.

We want to find the hypothesis test about the mean μ,

z=x¯-μ0σn=23-22415=0.97

Therefore, this is right tailed test with α=0.05, the critical value is given below,

za=z0.05=1.645

The critical region is z>z0.05. Here, z=0.97<z0.05=1.645

Hence, we reject H0at 5% level of significance as the value of zdoes not fall in the rejection region.

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Most popular questions from this chapter

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 16.7 months. One hundred randomly selected motor-vehicle-theft offenders in Sydney, Australia, had a mean length of imprisonment of 17.8 months. At the5% 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 6.0 months.

Refer to Exercise 9.18. 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 age at diagnosis of all people with early-onset dementia.

(d) is 55 years old.

(e) is less than 55 years old.

9.89 Job Gains and Losses. In the article "Business Employment Dynamics: New Data on Gross Job Gains and Losses" (Monthly Labor Review, Vol. 127. Issue 4. pp. 29-42). 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 20 quarters provided the net percentage gains (losses are negative gains) for jobs as presented on the WeissStats site. Use the technology of your choice to do the following.
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.
b. Obtain a normal probability plot, boxplot, histogram, and stem-and-leaf diagram of the data.
c. Remove the outliers (if any) from the data and then repeat part (a).
d. Comment on the advisability of using the z-test here.

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=24,σ=4,H0:μ=22,Ha:μ≠22

The following relationship holds between hypothesis tests and confidence intervals for one-mean t-procedures: For a two-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 μ0lies outside the 1-α-level confidence interval for μ. In each case, illustrate the preceding relationship by obtaining the appropriate one-mean t-interval and comparing the result to the conclusion of the hypothesis test in the specified exercise.

Part (a): Exercise 9.113

Part (b): Exercise 9.116

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