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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¯=24,n=15,σ=4,H0:μ=22,Ha:μ>22

Short Answer

Expert verified

The value of z is 1.94, critical value is 1.645,P=0.026and reject H0.

Step by step solution

01

Step 1. Given information.

Consider the given question,

x¯=24,n=15,σ=4,H0:μ=22,Ha:μ>22

02

Step 2. Consider the test hypothesis.

Consider the given hypothesis,

μis the population mean.

The test hypothesis is given below,

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=24-22415=1.94

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

role="math" localid="1651164260948" za=z0.05=1.645

The rejection region is z>z0.05.

Here, z=1.94>z0.05=1.645

Hence, we reject H0at 5% level of significance as the value of the test satisfice falls in the rejection region.

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Most popular questions from this chapter

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

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.

9.90 Hotels and Motels. The daily charges, in dollars, for a sample of 15 hotels and motels operating in South Carolina are provided on the WeissStats site. The data were found in the report South Caroline Statistical Abstract, sponsored by the South Carolina Budget and Control Board.
a. Use the one-mean z-test to decide, at the 5%significance level, whether the data provide sufficient evidence to conclude that the mean daily charge for hotels and motels operating in South Carolina is less than \(75. Assume a population standard deviation of \)22.40.

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.

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 α, the null hypothesis role="math" localid="1653038937481" 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 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

The normal probability curve and histogram 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.

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