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Chapter 10: Problem 90

Researchers interviewed street prostitutes in Canada and the United States. The mean age of the 100 Canadian prostitutes upon entering prostitution was 18 with a standard deviation of six. The mean age of the 130 United States prostitutes upon entering prostitution was 20 with a standard deviation of eight. Is the mean age of entering prostitution in Canada lower than the mean age in the United States? Test at a 1% significance level.

Short Answer

Expert verified
The mean age of entering prostitution in Canada is significantly lower than in the U.S. at the 1% significance level.

Step by step solution

01

Define the Hypotheses

We need to set up the null and alternative hypotheses to compare the means of the two independent samples.- Null Hypothesis \(H_0\): The mean age of entering prostitution in Canada is equal to that in the U.S., i.e., \(\mu_C = \mu_U\).- Alternative Hypothesis \(H_a\): The mean age of entering prostitution in Canada is less than that in the U.S., i.e., \(\mu_C < \mu_U\).
02

Determine the Test Statistic

We will use a two-sample Z-test for comparing the means of two independent samples, since we are comparing the means from two different groups with known standard deviations.The test statistic is given by:\[Z = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]where:- \(\bar{X}_1 = 18\), \(\sigma_1 = 6\), and \(n_1 = 100\) for Canadian prostitutes- \(\bar{X}_2 = 20\), \(\sigma_2 = 8\), and \(n_2 = 130\) for U.S. prostitutes
03

Calculate the Test Statistic

Substituting the given values into the test statistic formula, we have:\[Z = \frac{18 - 20}{\sqrt{\frac{6^2}{100} + \frac{8^2}{130}}} = \frac{-2}{\sqrt{0.36 + 0.4923}} = \frac{-2}{0.8566}\]\[Z \approx -2.336\]
04

Determine the Critical Value

Since this is a left-tailed test at a 1% significance level, we use a Z-table to find the critical value for \(\alpha = 0.01\). The critical Z-value is approximately -2.33.
05

Make a Decision

Compare the calculated Z-value with the critical value:- Calculated Z-value: -2.336- Critical Z-value: -2.33Since \(-2.336 < -2.33\), we reject the null hypothesis.
06

Conclusion

Based on our calculations and comparisons, we have sufficient evidence to conclude that the mean age of entering prostitution in Canada is significantly lower than that in the United States at a 1% significance level.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Two-sample Z-test
When dealing with two different groups and you want to compare their means, the two-sample Z-test is a useful tool. This test helps determine if there's a significant difference between the means of two independent populations.
In the given problem, the two-sample Z-test is applied to check if the average age of entering prostitution in Canada is genuinely lower than that in the United States.

The formula used in the two-sample Z-test is:
  • \( Z = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \)
Here, \( \bar{X}_1 \) and \( \bar{X}_2 \) are the sample means; \( \sigma_1 \) and \( \sigma_2 \) are the standard deviations; \( n_1 \) and \( n_2 \) represent the sizes of the two samples.
Using this formula ensures that we account for both differences in sample sizes and variations in data.
Defining the Significance Level
The significance level, often denoted by \( \alpha \), is a threshold set to decide when a result is statistically significant.
It quantifies the risk you鈥檙e willing to take in incorrectly rejecting a true null hypothesis.

In most cases, common significance levels are 0.05, 0.01, and 0.10. Lower values imply stricter criteria for rejecting the null hypothesis.
In this exercise, a 1% significance level is chosen, symbolized by \( \alpha = 0.01 \). This strict significance level means there's only a 1% risk of concluding that there is a difference in mean ages when there actually isn't.
The choice of significance level is crucial as it influences the sensitivity and specificity of your hypothesis test outcome.
Calculating the Test Statistic
The test statistic is a standardized value that helps in making a decision about the hypothesis.
In our two-sample Z-test, the test statistic is calculated based on the difference between the sample means relative to the variability in the original samples.

Using the information:
  • Mean age for Canadians \( \bar{X}_1 = 18 \)
  • Standard deviation \( \sigma_1 = 6 \)
  • Sample size \( n_1 = 100 \)
For U.S.:
  • Mean age \( \bar{X}_2 = 20 \)
  • Standard deviation \( \sigma_2 = 8 \)
  • Sample size \( n_2 = 130 \)
The Z statistic formula results in \( Z \approx -2.336 \), which indicates how many standard deviations the observed mean difference is away from the null hypothesis, where no difference exists.
Understanding the Critical Value
The critical value in hypothesis testing defines the cutoff point on the distribution of the test statistic.
This helps to determine whether to reject the null hypothesis.

In a left-tailed test at a 1% significance level, the critical value is a point on the Z-distribution below which only 1% of data falls.
In our exercise, this critical value is approximately -2.33.
We compare it to our calculated test statistic of -2.336.
  • If the test statistic is more extreme than the critical value, we reject the null hypothesis.
  • Here, since -2.336 is less than -2.33, it indicates strong evidence against the null hypothesis.
Critical values are vital as they ensure objectivity in decision-making during hypothesis testing.

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Most popular questions from this chapter

Use the following information to answer the next 12 exercises: The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites. Does it appear that the means are the same? Why or why not?

Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. 15 pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point. $$\begin{array}{|l|l|l|}\hline & {\text { Sample Mean Melting Temperatures }\left(^{\circ} F\right)} & {\text { Population Standard Deviation }} \\\ \hline \text { Alloy Gamma } & {800}&{95} \\ \hline \text { Alloy zeta } & {900} &{105} \\ \hline\end{array}$$ At the 1% significance level, what is your conclusion?

Use the following information to answer next five exercises. A study was conducted to test the effectiveness of a juggling class. Before the class started, six subjects juggled as many balls as they could at once. After the class, the same six subjects juggled as many balls as they could. The differences in the number of balls are calculated. The differences have a normal distribution. Test at the 1% significance level. $$ \begin{array}{|c|c|c|c|c|c|}\hline \text { Subject } & {\mathbf{A}} & {\mathbf{B}} & {\mathbf{c}} & {\mathbf{D}} & {\mathbf{E}} & {\mathbf{F}} \\\ \hline \text { Before } & {3} & {4} & {3} & {2} & {4} & {5} \\ \hline \text { After } & {4} & {5} & {6} & {4} & {5} & {7} \\ \hline\end{array} $$

Use the following information to answer the next five exercises. A study was conducted to test the effectiveness of a software patch in reducing system failures over a six-month period. Results for randomly selected installations are shown in Table 10.21. The 鈥渂efore鈥� value is matched to an 鈥渁fter鈥� value, and the differences are calculated. The differences have a normal distribution. Test at the 1% significance level. $$ \begin{array}{|l|l|l|l|l|l|l|}\hline \text { Installation } & {\mathbf{A}} & {\mathbf{B}} & {\mathbf{C}} & {\mathbf{D}} & {\mathbf{E}} & {\mathbf{F}} & {\mathbf{G}} & {\mathbf{H}} \\ \hline \text { Before } & {3} & {6} & {4} & {2} & {5} & {8} & {2} & {6} \\ \hline \text { After } & {1} & {5} & {2} & {0} & {1} & {0} & {2} & {2} \\ \hline\end{array} $$ State the null and alternative hypotheses.

One of the questions in a study of marital satisfaction of dual-career couples was to rate the statement 鈥淚鈥檓 pleased with the way we divide the responsibilities for childcare.鈥� The ratings went from one (strongly agree) to five (strongly disagree). Table 10.26 contains ten of the paired responses for husbands and wives. Conduct a hypothesis test to see if the mean difference in the husband鈥檚 versus the wife鈥檚 satisfaction level is negative (meaning that, within the partnership, the husband is happier than the wife). $$ \begin{array}{|l|l|l|l|l|l|l|}\hline \text { Wife's Score } & {2} & {2} & {3} & {3} & {4} & {2} & {1} & {1} & {2} \\ \hline \text { Husband's Score } & {2} & {2} & {1} & {3} & {2} & {1} & {1} & {1} & {2} \\ \hline\end{array} $$
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