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Chapter 9: Q. 38 (page 394)

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.

Short Answer

Expert verified

Neither test (z-test or t-test) would be used. But, some statistician would be used z-test because the population standard deviation is known.

Step by step solution

01

Step 1. Given Information 

02

Step 2. Consulting plots

From the given plots, it is observed that the distribution of the data is not approximately normally distributed.

03

Step 3. Explanation

Since the distribution of the data is not normally distributed and the sample size is small, neither test (z-test or t-test) would be used. But, some statistician would be used z-test because the population standard deviation is known.

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Most popular questions from this chapter

Ankle Brachial Index. The ankle brachial index (ABI) compares the blood pressure of a patient's arm to the blood pressure of the patient's leg. The ABI can be an indicator of different diseases, including arterial diseases. A healthy (or normal) ABI is 0.9 or greater. In a study by M. McDermott et al. titled "Sex Differences in Peripheral Arterial Disease: Leg Symptoms and Physical Functioning" (Journal of the American Geriatrics Society, Vol. 51, No. 2, Pp. 222-228), the researchers obtained the ABI of 187 women with peripheral arterial disease. The results were a mean ABI of 0.64 with a standard deviation of 0.15 At the 1 % significance level, do the data provide sufficient evidence to conclude that, on average, women with peripheral arterial disease have an unhealthy ABI?

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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.
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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 of-fenders in Sydney, Australia, had a mean length of imprisonment of 17.8 months. At the 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 is6months.

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d. Comment on the advisability of using the z-test here.

As we mentioned on page \(378\), the following relationship holds between hypothesis tests and confidence intervals for one mean \(z-\)procedures: For a two-tailed hypothesis test at the significance level \(\alpha\), the null hypothesis \(H_{0}:\mu =\mu_{0}\) will be rejected in favor of the alternative hypothesis \(H_{a}:\mu \neq \mu_{0}\) if and only if \(\mu_{0}\) lies outside the \((1-\infty)\) level confidence interval for \(\mu\). 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.

a. Exercise \(9.84\)

b. Exercise \(9.87\)
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