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A typical month, men spend \(\$ 178\) and women spend \(\$ 96\) on leisure activities," according to the results of an International Communications Research (ICR) for American Express poll, as reported in a USA Today Snapshot found on the Internet June 25,2005 Suppose random samples were taken from the population of male and female college students. Each student was asked to determine his or her expenditures for leisure activities in the prior month. The sample data results had a standard deviation of \(\$ 75\) for the men and a standard deviation of \(\$ 50\) for the women. a. If both samples were of size \(20,\) what is the standard error for the difference of two means? b. Assuming normality in leisure activity expenditures, is the difference found in the ICR poll significant at \(\alpha=0.05\) if the samples in part a are used? Explain.

Short Answer

Expert verified
Firstly, calculate the standard error for the difference of means using the given standard deviation and sample size. Secondly, find the difference of means using the provided mean values for men and women. Finally, perform a two-sample Z test to determine if the difference found in the ICR poll between male and female expenditures on leisure activities is significant at α=0.05.

Step by step solution

01

Calculation of Standard Error for Difference of Two Means

The formula for standard error (SE) when comparing two means is: \( SE = \sqrt{\frac{s1^2}{n1} + \frac{s2^2}{n2}} \), where 's' is the standard deviation and 'n' is the sample size. In this case, \(s1 = $ 75\) (men), \(s2 = $ 50\) (women), \(n1 = 20\) (men), and \(n2 = 20\) (women). Now, plug in the known values and compute the standard error: \( SE = \sqrt{\frac{75^2}{20} + \frac{50^2}{20}} \).
02

Calculation of the Difference of Means

The difference of means \(D\) is given by \(D = M1 - M2\), where \(M1\) and \(M2\) are the means of the two groups. In this case, \(M1 = $ 178\) (men), and \(M2 = $ 96\) (women). So, \(D = 178 - 96\).
03

Perform Hypothesis Testing

Assuming normality in leisure activity expenditures, conduct a two-sample Z test. The Null hypothesis \(H0\) is that there's no difference between men and women's spending on leisure i.e: \(D = 0\). The Alternative Hypothesis is \(Ha: D \neq 0\). Compute the Z statistic using the formula \(Z = \frac{D - H0}{SE}\), where \(D\) is the difference of means, \(H0\) is the hypothesized difference (in this case, 0) and \(SE\) is the standard error. Compare the computed Z-score with the critical Z-score for a two-tailed test at α=0.05. If the computed Z-score is more extreme than the critical Z-score, we reject the null hypothesis.

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Key Concepts

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

Standard Error
The standard error is a measure that tells us how much we can expect our sample mean to vary due to randomness. It provides a numerical estimate of the precision of the sample mean compared to the actual population mean. For two independent samples, the standard error of the difference in means is a special kind of standard error. When looking at two sets of data – for instance, men's and women's expenditures – we're interested in how accurately we can estimate the difference between their averages.

The formula to find the standard error for the difference between two means is: \[ SE = \sqrt{\frac{s1^2}{n1} + \frac{s2^2}{n2}} \] Here,
  • \( s1 \) and \( s2 \) are the standard deviations of each group.
  • \( n1 \) and \( n2 \) are the sample sizes of each group.
Take note that a smaller standard error indicates a more precise estimate of the population difference.
Two-Sample Z Test
A Two-Sample Z Test is a statistical method used to determine if there is a significant difference between the means of two groups. It is especially useful when the sample sizes are large, and the populations are normally distributed, or the sample sizes are equal and known to be normally distributed.

In this scenario, we are assessing whether the observed difference in leisure expenditures between men and women is statistically significant. We first calculate the Z statistic, which tells us how much the observed difference deviates from what we would expect if there was no true difference.

The formula for the Z statistic is: \[ Z = \frac{D - H0}{SE} \] Where:
  • \( D \) is the observed difference in sample means (Men's mean minus Women's mean).
  • \( H0 \) is the hypothesized mean difference (usually 0).
  • \( SE \) is the standard error obtained from our earlier formula.
If the Z statistic is further from zero than the critical Z value from the Z distribution table (for a given confidence level, such as 0.05), we can infer that there is indeed a significant difference.
Difference of Means
The concept of difference of means is straightforward, representing the quantitative difference between the average values of two groups. In our case, it's simply the difference in spending amounts on leisure activities between men and women.

To find the difference of means, we subtract the mean of the women's expenditures from the mean of the men's expenditures: \[ D = M1 - M2 \]Where:
  • \( M1 \) is the mean expenditure of men.
  • \( M2 \) is the mean expenditure of women.
This result reflects how much more, on average, one group spends compared to the other. It's a crucial step in hypothesis testing because it provides the observed value we compare against the expected (hypothesized) value.
Null Hypothesis
The null hypothesis is a fundamental concept in hypothesis testing. It essentially proposes that there is no effect or no difference, and it acts as a starting point for statistical testing. In our particular exercise, the null hypothesis states that there is no difference in leisure spending between men and women.

Formally, this could be presented as: \[ H0: D = 0 \]Where \( D \) is the difference of means. The null hypothesis suggests that any observed difference is due to sampling variability or random chance rather than a true difference. When performing a Z-test, we compare the calculated Z statistic to the standard normal distribution. If the Z score falls within the critical range (typically defined by the chosen significance level \( \alpha \)), we fail to reject the null hypothesis. If it falls outside, we reject the null hypothesis in favor of the alternative, suggesting a true difference in means exists. Thus, the null hypothesis plays a crucial role by providing a benchmark against which evidence from statistics is weighed.

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Most popular questions from this chapter

Two independent random samples of size 25 were taken from an English class and a chemistry class at a local community college. Students in both classes were asked to draw a 3 -inch line to the best of their ability without any measuring device (ruler, etc.). The following data resulted: $$\begin{array}{llll}\hline \text { English } & n=25 & \bar{x}=2.660 & s=0.617 \\\\\text { Chemistry } & \mathrm{n}=25 & \bar{x}=2.750 & \mathrm{s}=0.522 \\\\\hline\end{array}$$ At the 0.05 level of significance, is there a difference between the standard deviations for 3 -inch-line measurements from the English and chemistry classes?

A test that measures math anxiety was given to 50 male and 50 female students. The results were as follows: $$\begin{aligned}&\text { Males: } \quad \bar{x}=70.5, s=13.2\\\&\text { Females: } \bar{x}=75.7, s=13.6\end{aligned}$$ Construct a \(95 \%\) confidence interval for the difference between the mean anxiety scores.

Ten randomly selected college students who participated in a learning community were given pre-self-esteem and post-self-esteem surveys. A learning community is a group of students who take two or more courses together. Typically, each learning community has a theme, and the faculty involved coordinate assignments linking the courses. Research has shown that higher self-esteem, higher grade point averages (GPAs), and improved satisfaction in courses, as well as better retention rates, result from involvement in a learning community. The scores on the surveys are as follows:$$\begin{array}{lcccccccccc}\text { Student } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\\\\hline \text { Prescore } & 18 & 14 & 11 & 23 & 19 & 21 & 21 & 21 & 11 & 22 \\\\\text { Postscore } & 17 & 17 & 10 & 25 & 20 & 10 & 24 & 22 & 10 & 24 \\\\\hline\end{array}$$Does this sample of students show sufficient evidence that self-esteem scores were higher after participation in a learning community? Lower scores indicate higher self-esteem. Use the 0.05 level of significance and assume normality of score.

Find the value of \(t \star\) for the difference between two means based on an assumption of normality and thic information about two samples: $$\begin{array}{cccc}\text { Sample } & \text { Number } & \text { Mean } & \text { Std. Dev. } \\\\\hline 1 & 18 & 38.2 & 14.2 \\\2 & 25 & 43.1 & 10.6 \\\\\hline\end{array}$$

State the null hypothesis, \(H_{o},\) and the alternative hypothesis, \(H_{a},\) that would be used to test these claims: a. There is an increase in the mean difference between post-test and pre-test scores. b. Following a special training session, it is believed that the mean of the difference in performance scores will not be zero. c. On average, there is no difference between the readings from two inspectors on each of the selected parts. d. The mean of the differences between pre-self-esteem and post-self-esteem scores showed improvement after involvement in a college learning community.

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