# Test of Hypothesis with Significance Level Provided

9.56) 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 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 is 6.0 months.

9.68) What is the difference in assumptions between the one sample t-test and one sample t-test and the one-sample z-test?

Preliminary data analyses indicate that you can reasonably use a t-test to conduct each hypothesis tests required for exercise 9.70. Perform each t-test, using either the critical-value approach or the P-value approach. Comment on the practical significance of those tests whose results are statistically significant.

9.70) Previous studies have shown that urban bus drivers have an extremely stressful job, and a large proportion of drivers retire prematurely with disabilities due to occupational stress. These stresses come from a combination of physical and social sources such as traffic congestion, incessant time pressure, and unruly passengers. Among other variables, the researchers monitored diastolic blood pressure of bus drivers in downtown Stockholm, Sweden. The following data, in millimeters of mercury (mmHg), are based on the blood pressure obtained prior to intervention for the bus 41 bus drivers in the study.

95 77 99 81 85 81 84

77 79 100 83 89 73 79

80 70 90 74 79 76 65

91 88 83 90 93 69 89

95 80 70 66 90 94 93

77 81 75 95 73 63

At the 10% significance level, do the data provide sufficient evidence to conclude that the mean diastolic blood pressure of bus drivers in Stockholm, Sweden exceeds the normal diastolic blood pressure of 80 mm Hg? (Note: the sample mean and sample standard deviation of the data is 81.76 mm Hg, respectively.)

9.84) Suppose that you want to perform a hypothesis test for a population mean. Assume that the variable under consideration is normally distributed and that the population standard deviation is unknown.

a. Is it permissible to use the t-test to perform the hypothesis test? Explain your answer.

b. Is it permissible to use the one -sample Wilcoxon signed rank test to perform the hypothesis test? Explain your answer.

c. Which procedure is better to use, the t-test or the one sample Wilcoxon signed rank-test? Explain your answer.

For exercise 9.56, which employs the P-value approach to hypothesis testing, to perform those same hypothesis tests. In addition what is the strength of evidence against the null hypothesis.

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See the attached file where formatting is conserved.

For exercise 9.56, which employs the P-value approach to hypothesis testing, to perform those same hypothesis tests. In addition what is the strength of evidence against the null hypothesis.

9.56

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 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 is 6.0 months.

Null Hypothesis:Ho: M =16.7

Alternative Hypothesis:H1: M > 16.7

Mean=M = 16.7 months

Standard deviation =s= 6 months

sample size=n= 100

sx=standard error of mean=s/square root of n= 0.6 = ( 6 /square root of 100)

Significance level=alpha (a) = 5.00%

No of tails= 1

This is a 1 tailed test because we are testing sample mean>16.7

Since sample size= 100 >= 30 use normal distribution

sample mean= 17.8 months

Z at the 0.05 level of significance 1 tailed test = 1.6449

z=(sample mean-M )/sx= 1.8333 =(17.8-16.7)/0.6

Prob-value corresponding to z= 1.8333 is 3.34% which is less than 5.00%

Thus reject the null hypothesis

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

9.68

What is the difference in assumptions between the one sample t-test and one ...

#### Solution Summary

Questions on hypothesis testing have been answered.