An education organization claims that the mean SAT scores for male athletes and male non-athletes at a college are different. A random sample of 26 male athletes at the college has a mean SAT score of 1189 and a standard deviation of 218. A random sample of 18 male non-athletes at the college has a mean SAT score of 1376 and a standard deviation of 186. At α=0.05, can you support the organization’s claim? Interpret the decision in the context of the original claim. Assume the populations are normally distributed and the population variances are equal.
Table of contents
- 1. Intro to Stats and Collecting Data1h 14m
- 2. Describing Data with Tables and Graphs1h 56m
- 3. Describing Data Numerically2h 5m
- 4. Probability2h 17m
- 5. Binomial Distribution & Discrete Random Variables3h 6m
- 6. Normal Distribution and Continuous Random Variables2h 11m
- 7. Sampling Distributions & Confidence Intervals: Mean3h 23m
- Sampling Distribution of the Sample Mean and Central Limit Theorem19m
- Distribution of Sample Mean - ExcelBonus23m
- Introduction to Confidence Intervals15m
- Confidence Intervals for Population Mean1h 18m
- Determining the Minimum Sample Size Required12m
- Finding Probabilities and T Critical Values - ExcelBonus28m
- Confidence Intervals for Population Means - ExcelBonus25m
- 8. Sampling Distributions & Confidence Intervals: Proportion2h 10m
- 9. Hypothesis Testing for One Sample5h 8m
- Steps in Hypothesis Testing1h 6m
- Performing Hypothesis Tests: Means1h 4m
- Hypothesis Testing: Means - ExcelBonus42m
- Performing Hypothesis Tests: Proportions37m
- Hypothesis Testing: Proportions - ExcelBonus27m
- Performing Hypothesis Tests: Variance12m
- Critical Values and Rejection Regions28m
- Link Between Confidence Intervals and Hypothesis Testing12m
- Type I & Type II Errors16m
- 10. Hypothesis Testing for Two Samples5h 37m
- Two Proportions1h 13m
- Two Proportions Hypothesis Test - ExcelBonus28m
- Two Means - Unknown, Unequal Variance1h 3m
- Two Means - Unknown Variances Hypothesis Test - ExcelBonus12m
- Two Means - Unknown, Equal Variance15m
- Two Means - Unknown, Equal Variances Hypothesis Test - ExcelBonus9m
- Two Means - Known Variance12m
- Two Means - Sigma Known Hypothesis Test - ExcelBonus21m
- Two Means - Matched Pairs (Dependent Samples)42m
- Matched Pairs Hypothesis Test - ExcelBonus12m
- Two Variances and F Distribution29m
- Two Variances - Graphing CalculatorBonus16m
- 11. Correlation1h 24m
- 12. Regression3h 33m
- Linear Regression & Least Squares Method26m
- Residuals12m
- Coefficient of Determination12m
- Regression Line Equation and Coefficient of Determination - ExcelBonus8m
- Finding Residuals and Creating Residual Plots - ExcelBonus11m
- Inferences for Slope31m
- Enabling Data Analysis ToolpakBonus1m
- Regression Readout of the Data Analysis Toolpak - ExcelBonus21m
- Prediction Intervals13m
- Prediction Intervals - ExcelBonus19m
- Multiple Regression - ExcelBonus29m
- Quadratic Regression15m
- Quadratic Regression - ExcelBonus10m
- 13. Chi-Square Tests & Goodness of Fit2h 21m
- 14. ANOVA2h 29m
10. Hypothesis Testing for Two Samples
Two Means - Unknown, Equal Variance
Problem 8.RE.15
Textbook Question
In Exercises 11–16, test the claim about the difference between two population means μ1 and μ2 at the level of significance α. Assume the samples are random and independent, and the populations are normally distributed.
Claim: μ1≠ μ2; α=0.01. Assume (σ1)^2 = (σ2)^2
Sample statistics: x̅1= 61, s1= 3.3, n1= 5 and x̅2= 55.1, s2= 1.2, n2= 7
Verified step by step guidance1
Identify the null hypothesis \( H_0 \) and the alternative hypothesis \( H_a \) based on the claim. Since the claim is \( \mu_1 \neq \mu_2 \), set \( H_0: \mu_1 = \mu_2 \) and \( H_a: \mu_1 \neq \mu_2 \).
Since the population variances are assumed equal (\( \sigma_1^2 = \sigma_2^2 \)), use the pooled variance formula to estimate the common variance. Calculate the pooled variance \( s_p^2 \) as:
\[
s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}
\]
Calculate the test statistic \( t \) using the formula for two independent samples with equal variances:
\[
t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}
\]
where \( s_p = \sqrt{s_p^2} \).
Determine the degrees of freedom for the test, which is \( df = n_1 + n_2 - 2 \).
Find the critical value(s) from the \( t \)-distribution table for a two-tailed test at significance level \( \alpha = 0.01 \) with \( df \) degrees of freedom. Compare the calculated \( t \)-statistic to the critical value(s) to decide whether to reject or fail to reject the null hypothesis.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Hypothesis Testing for Two Population Means
This involves testing whether there is a statistically significant difference between the means of two populations. The null hypothesis typically states that the means are equal (μ1 = μ2), while the alternative reflects the claim (μ1 ≠ μ2). The test uses sample data to decide whether to reject the null hypothesis at a given significance level.
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Pooled Variance and Equal Population Variances Assumption
When the population variances are assumed equal (σ1² = σ2²), a pooled variance estimate combines the sample variances to improve the accuracy of the test statistic. This pooled variance is a weighted average of the sample variances, accounting for different sample sizes, and is used in the calculation of the t-test statistic.
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Level of Significance (α) and Critical Values
The level of significance, α, is the probability of rejecting the null hypothesis when it is true (Type I error). For a two-tailed test with α = 0.01, critical values define the rejection regions in the t-distribution. If the test statistic falls beyond these critical values, the null hypothesis is rejected in favor of the alternative.
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