Question

Ages of Gamblers The mean age of a random sample of 25 people who were playing the slot machines is 48.7 years, and the standard deviation is 6.8 years. The mean age of a random sample of 35 people who were playing roulette is 55.3 with a standard deviation of 3.2 years. Can it be concluded at \(\alpha=0.05\) that the mean age of those playing the slot machines is less than those playing roulette?

Short answer

Yes, there is sufficient evidence to conclude the mean age of slot players is less than roulette players at \( \alpha = 0.05 \).

Step-by-step solution

Identify Hypotheses

Begin by stating the null and alternative hypotheses. The null hypothesis (H_0) states that the mean age of slot machine players is equal to the mean age of roulette players, i.e., \( \mu_1 = \mu_2 \). The alternative hypothesis (H_a) states that the mean age of slot machine players is less than the mean age of roulette players, i.e., \( \mu_1 < \mu_2 \).

Set Significance Level

The significance level, \( \alpha \), provided in the problem is 0.05. This level of significance indicates the probability of rejecting the null hypothesis when it is actually true.

Calculate the Test Statistic

To find the test statistic, use the formula for the difference between two means: \[ z = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]Here, \( \bar{x}_1 = 48.7 \), \( \bar{x}_2 = 55.3 \), \( s_1 = 6.8 \), \( s_2 = 3.2 \), \( n_1 = 25 \), \( n_2 = 35 \). Calculate:\[ z = \frac{(48.7 - 55.3)}{\sqrt{\frac{6.8^2}{25} + \frac{3.2^2}{35}}} = \frac{-6.6}{\sqrt{1.8496 + 0.2928}} = \frac{-6.6}{\sqrt{2.1424}} = \frac{-6.6}{1.4637} \approx -4.51 \]

Determine Critical Value

Since this is a one-tailed test with \( \alpha = 0.05 \), the critical value for \( z \) from the standard normal distribution table is approximately -1.645. This means that the null hypothesis will be rejected if \( z < -1.645 \).

Make a Decision

Compare the calculated test statistic \( z \approx -4.51 \) to the critical value \(-1.645\). Since \(-4.51 < -1.645\), we reject the null hypothesis.

Conclusion

Based on the test, there is sufficient evidence at the \( \alpha = 0.05 \) level to conclude that the mean age of those playing slot machines is less than those playing roulette.

Key concepts

Null Hypothesis

The null hypothesis is a fundamental concept in statistical hypothesis testing. It's denoted as ?`H_0`? and is essentially a statement of no effect or no difference. In our example, the null hypothesis asserts that the mean age of slot machine players is equal to the mean age of roulette players. Mathematically, we write this as \( \mu_1 = \mu_2 \). In hypothesis testing, the null hypothesis is the statement that is initially assumed to be true. The objective is to determine whether there is enough evidence from the sample data to reject this hypothesis in favor of the alternative. Remember, the null hypothesis is not proven to be true but is assumed true until evidence suggests otherwise.Understanding the null hypothesis helps set the stage for further analysis and is crucial when making conclusions about data. It's our starting point in any statistical test, providing a baseline or standard for comparison.

Alternative Hypothesis

The alternative hypothesis, denoted by \( H_a \), is what researchers aim to support. It opposes the null hypothesis. In the context of our example, the alternative hypothesis suggests that the mean age of slot machine players is less than that of roulette players. This is expressed mathematically as \( \mu_1 < \mu_2 \).A key part of the alternative hypothesis is its specific direction. In our test, \( H_a \) is a one-tailed hypothesis because we are only interested in finding evidence that one mean is less than the other. In contrast, a two-tailed hypothesis checks for any difference, either greater or less.Selecting the right alternative hypothesis is crucial. It directly affects the statistical method we use and the interpretation of test results. Always choose it based on the research question and what you wish to prove.

Z-Test

The z-test is a statistical method used to determine if there is a significant difference between the means of two groups. It assesses whether the observed data are at odds with the null hypothesis. For comparison of means, especially with large sample sizes, the z-test is highly suitable.In this context, we use a formula to calculate the z-test statistic for two independent means: \[ z = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]Here, \( \bar{x}_1 \) and \( \bar{x}_2 \) are sample means, \( s_1 \) and \( s_2 \) are standard deviations, and \( n_1 \) and \( n_2 \) are sample sizes. The z-test provides a z-score, which conveys how many standard deviations away a data point is from the mean. A benefit of the z-test is its reliance on the normal distribution, allowing us to determine probabilities and make data-driven decisions based on well-established statistical principles. However, it's most accurate when sample sizes are large (n > 30).

Significance Level

The significance level, symbolized by \( \alpha \), represents the probability of rejecting the null hypothesis when it is actually true. In layman's terms, it measures the risk we're willing to take for making a Type I error, which occurs when we mistakenly find a "statistical effect" that isn't there.For the given problem, \( \alpha = 0.05 \). This means there is a 5% risk that we'll incorrectly reject the null hypothesis if it is true. This level is a common choice in many studies, reflecting a balance between being too lenient and too stringent.The chosen significance level is crucial because it affects the critical value, which is the threshold the test statistic must exceed in order to reject the null hypothesis. For a \( \alpha = 0.05 \) and a one-tailed test, you would look for a critical z-value of approximately -1.645. Selecting your significance level thoughtfully is also a part of ensuring your statistical test is both accurate and meaningful.