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Test \(H_{0}: \mu_{A}=\mu_{B}\) vs \(H_{a}: \mu_{A} \neq \mu_{B}\) using the fact that Group A has 8 cases with a mean of 125 and a standard deviation of 18 while Group \(\mathrm{B}\) has 15 cases with a mean of 118 and a standard deviation of 14

Short Answer

Expert verified
The t-statistic calculated was 1.26 which is less than the critical value of 2.08 at a 5% significance level with 21 degrees of freedom. Therefore, there is insufficient evidence to reject the null hypothesis that \(\mu_A = \mu_B\), meaning it cannot be concluded that the means of groups A and B are significantly different.

Step by step solution

01

Calculate Pooled Variance

First, calculate the weighted average of the variances (or pooled variance). The formula is: \((n_A-1)s_A^2 + (n_B-1)s_B^2)\) / \((n_A+n_B-2)\). Where \(n_i\) is the sample size of group i, and \(s_i\) is the standard deviation of group i. Substituting in the given numbers, we get: \((7 * (18)^2 + 14*(14)^2)/(8+15-2) = 254.22\)
02

Calculate t-Statistic

Now that we have the pooled variance, we can calculate the t-statistic. The formula is: \((M_A-M_B)\) / \( \sqrt{(PooledVariance/n_{A}) + (PooledVariance/n_{B})}\). Substituting our values gives us: \((125 - 118) / \sqrt{(254.22/8) + (254.22/15)} = 1.26\)
03

Calculate Degrees of Freedom

For a two-sample t-test, the degrees of freedom (df) is typically calculated as the sum of the sample sizes of the two groups, minus 2. In this case, df = \(8 + 15 - 2 = 21\)
04

Interpret the Test Statistic

Using a t-table or relevant software, one can look up the probability of obtaining a t-statistic as extreme as the one calculated (absolute value) under the null hypothesis. For a two-tailed test at 5% significance level and 21 degrees of freedom, the critical value is roughly 2.08. Since 1.26 is less than 2.08, we fail to reject the null hypothesis: there isn't enough statistical evidence to say that the two population means are different.

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Key Concepts

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

Pooled Variance
Pooled variance is a method used to estimate the variance of two populations that are assumed to have the same variance. In a two-sample t-test, it's crucial to combine the variances of both samples into a single, pooled estimate. This pooled variance represents the weighted average of the variances from both groups.

To calculate it, we use the formula \[ (n_A - 1)s_A^2 + (n_B - 1)s_B^2 \/ (n_A + n_B - 2) \], where \(n_A\) and \(n_B\) are the sample sizes and \(s_A^2\) and \(s_B^2\) are the sample variances. The degrees of freedom for each sample, \(n_A - 1\) and \(n_B - 1\), ensure that the variance is appropriately weighted for each group's contribution.

For the given problem, substituting the numbers into the formula yielded a pooled variance of 254.22. This step is foundational, as the pooled variance is used in computing the t-statistic, which ultimately helps us determine if there's a significant difference between the groups' means.
T-Statistic Calculation
Next comes the t-statistic calculation, which is the centerpiece of the two-sample t-test. The t-statistic reflects how much the sample means differ from each other relative to the variation within the samples. To compute the t-statistic, we apply the formula \[ (\bar{X}_A - \bar{X}_B) \/ \sqrt{(PooledVariance/n_A) + (PooledVariance/n_B)} \], where \(\bar{X}_A\) and \(\bar{X}_B\) are the sample means.

What's happening here is we're scaling the difference between the sample means by the standard error, which spreads the variance across the combined sample sizes. This calculation tells us the number of standard errors away the sample means are from one another. With the values given, the t-statistic turned out to be 1.26, a measure that we will use to assess the significance of the difference observed.
Degrees of Freedom
Degrees of freedom is a concept tied to the reliability of a statistical estimate. It's roughly the number of independent pieces of information that go into the estimate. For a two-sample t-test, the total degrees of freedom are calculated by adding the sample sizes of both groups, then subtracting two (the number of sample means estimated).

The formula is \(df = n_A + n_B - 2\). Degrees of freedom are crucial when referring to the t-distribution, as they shape the distribution. More degrees of freedom typically mean the distribution is closer to the normal distribution. In our exercise, with 8 participants in group A and 15 in group B, we have a total of 21 degrees of freedom. This figure is intrinsic to the final steps of hypothesis testing, where we compare our t-statistic against critical values from the t-distribution.
Hypothesis Testing
Finally, hypothesis testing is a systematic method to decide whether to accept or reject a hypothesis about a population parameter, based on sample data. In this case, we're testing the null hypothesis \(H_0: \mu_A = \mu_B\) against the alternative hypothesis \(H_a: \mu_A eq \mu_B\).

We compare the calculated t-statistic to a critical value from the t-distribution that corresponds to the degrees of freedom from our test and the desired level of significance. If our t-statistic exceeds the critical value, we reject the null hypothesis, indicating a statistically significant difference between the groups.

Using a t-table or software, we find the critical value for a two-tailed test with 21 degrees of freedom at a 5% significance level is approximately 2.08. Since our calculated t-statistic of 1.26 is less than this critical value, we do not reject the null hypothesis. Therefore, there isn't sufficient evidence to conclude that there is a significant difference between the means of Group A and Group B.

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Most popular questions from this chapter

In Exercises 6.153 to \(6.158,\) if random samples of the given sizes are drawn from populations with the given proportions: (a) Find the mean and standard error of the distribution of differences in sample proportions, \(\hat{p}_{A}-\hat{p}_{B}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 50 from population \(A\) with proportion 0.70 and samples of size 75 from population \(B\) with proportion 0.60

In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion: (a) Find the mean and standard error of the distribution of sample proportions. (b) If the sample size is large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 1000 from a population with proportion 0.70

Standard Error from a Formula and a Bootstrap Distribution In Exercises 6.19 to \(6.22,\) use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of home team wins in soccer, with \(n=120\) and \(\hat{p}=0.583\)

Random samples of the given sizes are drawn from populations with the given means and standard deviations. For each scenario: (a) Find the mean and standard error of the distribution of differences in sample means, \(\bar{x}_{1}-\bar{x}_{2}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 25 from Population 1 with mean 6.2 and standard deviation 3.7 and samples of size 40 from Population 2 with mean 8.1 and standard deviation 7.6

How big is the home field advantage in the National Football League (NFL)? In Exercise 6.240 on page 419 , we examine a difference in means between home and away teams using two separate samples of 80 games from each group. However, many factors impact individual games, such as weather conditions and the scoring of the opponent. It makes more sense to investigate this question using a matched pairs design, using scores for home and away teams matched for the same game. The data in NFLScores2011 include the points scored by the home and away team in 256 regular season games in \(2011 .\) We will treat these games as a sample of all NFL games. Estimate average home field scoring advantage and find a \(90 \%\) confidence interval for the mean difference.

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