Problem 14 - Basic Statistics for Business and Economics with Student

Question

(a) state the decision rule, (b) compute the pooled estimate of the population variance, (c) compute the test statistic, (d) state your decision about the null hypothesis, and (e) estimate the \(p\) -value. The null and alternate hypotheses are: $$ \begin{array}{l} H_{0}: \mu_{1}=\mu_{2} \\ H_{1}: \mu_{1} \neq \mu_{2} \end{array} $$ A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of \(12 .\) A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard deviation of \(15 .\) At the .10 significance level, is there a difference in the population means?

Step-by-Step Solution

Answer

Fail to reject the null hypothesis; insufficient evidence to suggest a difference at 0.10 significance level.

Step 1: State the Decision Rule

To determine if there is a significant difference between the population means at a 0.10 significance level, we will use a two-tailed t-test for independent samples. The decision rule is to reject the null hypothesis if the calculated t-statistic falls beyond the critical t-value from the t-distribution table for the given degrees of freedom. For a two-tailed test at 0.10 significance level, the critical t-value is determined by the degrees of freedom.

Step 2: Compute the Pooled Estimate of Population Variance

The pooled estimate of the population variance is computed using the formula:\[\text{Pooled Variance} = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}\]Here, \(n_1 = 15\), \(s_1 = 12\), \(n_2 = 17\), and \(s_2 = 15\). Plug in these values:\[\text{Pooled Variance} = \frac{(15 - 1) \cdot 12^2 + (17 - 1) \cdot 15^2}{15 + 17 - 2} = \frac{14 \cdot 144 + 16 \cdot 225}{30} = \frac{2016 + 3600}{30} = \frac{5616}{30} = 187.2\]

Step 3: Compute the Test Statistic

The t-statistic for two independent samples is given by:\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\left(\frac{1}{n_1} + \frac{1}{n_2}\right) s_p^2}}\]Where \(\bar{x}_1 = 350\), \(\bar{x}_2 = 342\), and \(s_p^2 = 187.2\):\[t = \frac{350 - 342}{\sqrt{\left(\frac{1}{15} + \frac{1}{17}\right) \cdot 187.2}} = \frac{8}{\sqrt{(0.0667 + 0.0588) \cdot 187.2}} = \frac{8}{\sqrt{0.1255 \cdot 187.2}} = \frac{8}{4.84} \approx 1.65\]

Step 4: State the Decision about the Null Hypothesis

The degrees of freedom for the test are \(n_1 + n_2 - 2 = 30\). Using a t-table, the critical t-value at 0.10 significance level for a two-tailed test with 30 degrees of freedom is approximately 1.697. Since the calculated t-statistic (1.65) is less than the critical t-value (1.697), we fail to reject the null hypothesis. Thus, there is no sufficient evidence at the 0.10 significance level to suggest a difference in population means.

Step 5: Estimate the P-value

To find the p-value, we compare the calculated t-statistic to a t-distribution table or use a statistical software. The p-value corresponding to a t-statistic of 1.65 with 30 degrees of freedom is approximately 0.108 (two-tailed). Because the p-value (0.108) is greater than the significance level (0.10), the result is not statistically significant.

Key Concepts

Pooled Variance

Pooled variance is a crucial part of hypothesis testing when dealing with two independent samples. It enables us to combine the variability from two different groups into a single measure, providing a clearer picture of the overall variation. Calculating pooled variance involves using sample sizes and variances from both groups involved in the analysis. To find it, use the formula: \[ \text{Pooled Variance} = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} \] Here, \(n_1\) and \(n_2\) are the sample sizes, while \(s_1\) and \(s_2\) are the sample standard deviations for the two groups respectively.

• It reflects a weighted average of the variance from each group.
• This method assumes that the variances of the two populations are equal.
• Pooled variance is critical for calculating the t-statistic, which we will explore next.

By unifying the data from both samples into one metric, you obtain a more reliable measure of variation, making the t-test results more robust.

t-test

The t-test is a widely used statistical method that helps in determining whether there is a meaningful difference between the means of two groups. Specifically, in this exercise, we use the t-test to compare the mean of two independent samples to see if they essentially come from the same population.The formula for the test statistic in an independent samples t-test is:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\left(\frac{1}{n_1} + \frac{1}{n_2}\right) s_p^2}} \]Where:

• \(\bar{x}_1\) and \(\bar{x}_2\) are sample means.
• \(n_1\) and \(n_2\) are the sample sizes.
• \(s_p^2\) is the pooled variance.

By examining the calculated t-value against a critical t-value or by evaluating the associated p-value, you can determine if the difference in means is statistically significant or simply due to sampling variability.

Significance Level

The significance level, often denoted as \( \alpha \), is a threshold we set before conducting hypothesis tests. It represents our willingness to accept a false rejection of the null hypothesis (Type I error). Common significance levels are 0.05, 0.01, and 0.10, with smaller values indicating stricter criteria for rejecting the null hypothesis.For example, in this case, the chosen significance level is 0.10. This means there is a 10% risk of concluding that a difference exists when there actually isn't one.

• A higher significance level like 0.10 allows for a larger margin of error, making it easier to reject the null hypothesis.
• It's crucial to decide on this level before performing the test to maintain the integrity of the results.
• The chosen significance level directly influences the determination of the critical t-value, which is crucial for making decisions about the null hypothesis.

Critical t-value

In hypothesis testing, the critical t-value acts as a boundary that helps us decide whether to reject the null hypothesis. It's determined by the chosen significance level and the degrees of freedom, which in this case is based on the sample sizes.For this exercise, with a 0.10 significance level and degrees of freedom calculated as \( n_1 + n_2 - 2 \), you find the critical t-value using a t-distribution table.

• If the calculated t-statistic is greater than the critical t-value (for a right or two-tailed test), we reject the null hypothesis.
• If it is less (or more negative in a left-tailed test), the null hypothesis remains unchallenged.
• Knowing the critical t-value allows us to establish a decision rule that guides the analysis outcome.

Thus, the critical t-value is a pivotal element for identifying significant differences and deciding the fate of the null hypothesis.