Math Problem Statement
Conduct the stated hypothesis test for μ 1− μ 2. μ 1− μ 2. Assume that the samples are independent and randomly selected from normal populations with equal population variances ( σ 12= σ 22)( σ 12= σ 22).
H0 : μ 1− μ 2=0H0 : μ 1− μ 2=0
H1 : μ 1− μ 2 < 0H1 : μ 1− μ 2 < 0
α =0.005 α =0.005
n1=29n1=29
x̄ 1=8.51 x̄ 1=8.51
s1=1.26s1=1.26
n2=23n2=23
x̄ 2=9.36 x̄ 2=9.36
s2=1.35s2=1.35
T-Distribution Table
a. Calculate the test statistic.
t=t=
Round to three decimal places if necessary
b. Determine the critical value(s) for the hypothesis test.
Round to three decimal places if necessary
c. Conclude whether to reject the null hypothesis or not based on the test statistic.
Reject
Fail to Reject
Solution
We are conducting a hypothesis test to compare the difference between two population means, and , with the following information:
Hypotheses:
- Null hypothesis (): (no difference in population means)
- Alternative hypothesis (): (mean of population 1 is less than mean of population 2)
Given Information:
- Significance level:
- Sample 1: , ,
- Sample 2: , ,
- Population variances are assumed equal:
Steps to solve the problem:
a. Calculate the test statistic.
Since the population variances are assumed to be equal, we will use the pooled variance and a two-sample t-test formula. The test statistic is calculated as:
Where:
- and are the sample means,
- and are the sample sizes,
- is the pooled standard deviation, calculated as:
Let’s compute the test statistic.
b. Determine the critical value.
To determine the critical value for a one-tailed t-test at a significance level , we need the degrees of freedom (df) and the t-distribution table.
c. Conclude whether to reject the null hypothesis.
We will compare the test statistic calculated in part (a) to the critical value from part (b). If the test statistic is less than the critical value (in the negative direction), we will reject the null hypothesis. Otherwise, we fail to reject it.
Let's compute the values step by step.### Results:
a. Test Statistic:
b. Critical Value: For a one-tailed test with and 50 degrees of freedom:
c. Conclusion: Since the test statistic () is greater than the critical value (), we fail to reject the null hypothesis. This means there is not enough evidence at the level to conclude that .
Would you like further details or explanations?
Here are 5 related questions for deeper exploration:
- How would the result change if the significance level were 0.01?
- How do we perform a two-tailed test in this scenario?
- What is the role of the pooled variance in hypothesis testing?
- How would unequal population variances affect the test?
- How can the sample sizes influence the test statistic and outcome?
Tip: The degrees of freedom in a t-test affect the shape of the t-distribution and hence the critical value. Always check if it's correct for your sample sizes.
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Math Problem Analysis
Mathematical Concepts
Hypothesis Testing
Two-Sample T-Test
Pooled Variance
Formulas
Test Statistic t = (x̄1 - x̄2) / sp √(1/n1 + 1/n2)
Pooled Standard Deviation sp = √[((n1 - 1)s1^2 + (n2 - 1)s2^2) / (n1 + n2 - 2)]
Theorems
Central Limit Theorem
T-Distribution
Suitable Grade Level
Undergraduate Level - Statistics
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