Question

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, the mean age of slot machine players is significantly less than that of roulette players at \( \alpha = 0.05 \).

Step-by-step solution

Define Hypotheses

We start by defining the null and alternative hypotheses for the problem.\\( H_0: \mu_1 = \mu_2 \) (The mean ages are equal)\\( H_a: \mu_1 < \mu_2 \) (The mean age of slot machine players is less than that of roulette players)

Formulate the Test Statistic

Since the standard deviations are given for independent samples, we use a two-sample t-test for the means. The test statistic is given by:\\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] \where \( \bar{x}_1 = 48.7 \), \( s_1 = 6.8 \), \( n_1 = 25 \) for slot machines, and \( \bar{x}_2 = 55.3 \), \( s_2 = 3.2 \), \( n_2 = 35 \) for roulette.

Calculate the Test Statistic

Substituting the given values into the formula, we proceed to calculate the test statistic:\\[ t = \frac{48.7 - 55.3}{\sqrt{\frac{6.8^2}{25} + \frac{3.2^2}{35}}} = \frac{-6.6}{\sqrt{\frac{46.24}{25} + \frac{10.24}{35}}} \] \\[ = \frac{-6.6}{\sqrt{1.8496 + 0.2926}} = \frac{-6.6}{\sqrt{2.1422}} \] \\[ = \frac{-6.6}{1.4636} \approx -4.51 \]

Determine the Critical Value

Since the claim is that the mean age of slot machine players is **less than** that of roulette players, we use a one-tailed t-test. The degrees of freedom can be approximated using the smaller of \( n_1 - 1 \) and \( n_2 - 1 \), so \( \text{df} = 24 \).\Using a t-distribution table or calculator and \( \alpha = 0.05 \), we find the critical value \( t_c = -1.711 \).

Make a Decision

Compare the calculated test statistic to the critical value. If \( t \lt t_c \), we reject \( H_0 \). In this case, \(-4.51 \lt -1.711\), so we reject \( H_0 \).

Conclusion

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

Key concepts

Hypothesis Testing

Hypothesis testing is a method used to make decisions about a population based on sample data. In the context of comparing two groups, it begins with formulating a null hypothesis (\(H_0\)) and an alternative hypothesis (\(H_a\)).

• Null Hypothesis: This is a statement that there is no effect or no difference, in this case, suggesting that the mean ages of two groups playing different games are equal (\(H_0: \mu_1 = \mu_2\)).
• Alternative Hypothesis: This indicates the presence of an effect or a difference. Specifically, that the mean age of slot machine players is less than that of roulette players (\(H_a: \mu_1 < \mu_2\)).

Hypothesis testing relies on evidence to either reject the null hypothesis or fail to reject it. It doesn’t "prove" the alternative hypothesis but only indicates whether the evidence is sufficient to support the claim.

Standard Deviation

Standard deviation measures the dispersion or variability in a set of data. In simple terms, it shows how much individual data points deviate from the mean (average) of the dataset. A smaller standard deviation means data points are closer to the mean, and a larger standard deviation indicates more spread out data.

• Standard Deviation of Slot Players: Here, the standard deviation (\(s_1 = 6.8\)) implies how much the ages of slot machine players vary from their mean age of 48.7 years.
• Standard Deviation of Roulette Players: Similarly, the roulette players have a smaller deviation (\(s_2 = 3.2\)), showing less age variability around the mean age of 55.3 years.

Understanding standard deviation helps assess the degree of overlap or divergence between two datasets, which is crucial in a two-sample t-test.

Critical Value

The critical value in hypothesis testing is a threshold that the test statistic must exceed to reject the null hypothesis. It is determined by the chosen significance level (also known as alpha, \(\alpha\)) and the degrees of freedom of the data.To find the critical value:

• The level of significance (\(\alpha = 0.05\)) is used, which corresponds to a 5% risk of concluding a difference exists when there is none.
• For this test, a one-tailed test is conducted because the alternative hypothesis specifies a direction (less than). One may present results by consulting a t-distribution table or using statistical software.
• The degrees of freedom for this sample size are approximated to the smaller of the two, resulting in \(df = 24\), giving a critical value of \(t_c = -1.711\).

When the calculated test statistic is more extreme than the critical value, it supports rejecting the null hypothesis.

Test Statistic

The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It allows us to determine the likelihood of observing the sample data under the null hypothesis.

• Formula: For a two-sample t-test, it is computed as follows: \\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]where \(\bar{x}_1\), \(\bar{x}_2\) are sample means, \(s_1\), \(s_2\) are standard deviations, and \(n_1\), \(n_2\) are sample sizes.
• Calculation: Substituting in the given values results in a test statistic of \(t \approx -4.51\). This value tells us how many standard deviations the observed difference in sample means is from zero (the null hypothesis).

The test statistic is compared against the critical value to decide whether to reject the null hypothesis.

Significance Level

The significance level, denoted by \(\alpha\), represents the probability of rejecting the null hypothesis when it is actually true, known as a Type I error. It is a threshold set before conducting a hypothesis test to decide how extreme a test statistic must be for the null hypothesis to be rejected.

• Selection of \(\alpha\): In this test, a significance level of 0.05 is used. This means there is a 5% chance of incorrectly concluding that there is a difference in mean ages between the two groups when none actually exists.
• Impact on Decision: The lower the significance level, the stricter the criteria for rejecting the null hypothesis. In practical terms, a 0.05 level implies moderate evidence is required to support the claim that the mean age of slot machine players is less than that of roulette players.

Choosing the right significance level is crucial since it balances the risks of errors against the practical consequences of those decisions.