Math Problem Statement

You wish to test the following claim ( H a ) at a significance level of
α

0.01 .

H o : μ 1

μ 2

H a : μ 1

μ 2

You believe both populations are normally distributed, but you do not know the standard deviations for either. However, you also have no reason to believe the variances of the two populations are not equal. You obtain a sample of size
n 1

12 with a mean of
M 1

56.6 and a standard deviation of
S D 1

12.6 from the first population. You obtain a sample of size
n 2

18 with a mean of
M 2

52.6 and a standard deviation of
S D 2

14.2 from the second population.

What is the p-value for this test? For this calculation, use the conservative under-estimate for the degrees of freedom as mentioned in the textbook. (Report answer accurate to four decimal places.) p-value =

Solution

To calculate the p-value for this test, we will follow these steps:

Step 1: Define Hypotheses

  • Null Hypothesis (H₀): μ1=μ2\mu_1 = \mu_2
  • Alternative Hypothesis (Hₐ): μ1>μ2\mu_1 > \mu_2

This is a one-tailed test since the alternative hypothesis suggests that μ1\mu_1 is greater than μ2\mu_2.

Step 2: Check Assumptions

  • Populations are normally distributed.
  • Population variances are assumed to be equal but unknown.

Step 3: Pooled Standard Deviation

Since the population variances are assumed to be equal, we calculate the pooled standard deviation SpS_p:

Sp=(n11)SD12+(n21)SD22n1+n22S_p = \sqrt{\frac{(n_1 - 1) \cdot SD_1^2 + (n_2 - 1) \cdot SD_2^2}{n_1 + n_2 - 2}}

Given:

  • n1=12n_1 = 12, SD1=12.6SD_1 = 12.6
  • n2=18n_2 = 18, SD2=14.2SD_2 = 14.2

Sp=(121)12.62+(181)14.2212+182S_p = \sqrt{\frac{(12 - 1) \cdot 12.6^2 + (18 - 1) \cdot 14.2^2}{12 + 18 - 2}} Sp=11158.76+17201.6428S_p = \sqrt{\frac{11 \cdot 158.76 + 17 \cdot 201.64}{28}} Sp=1746.36+3427.8828=5174.2428=184.0813.57S_p = \sqrt{\frac{1746.36 + 3427.88}{28}} = \sqrt{\frac{5174.24}{28}} = \sqrt{184.08} \approx 13.57

Step 4: Test Statistic (t-value)

The test statistic for a two-sample t-test is:

t=M1M2Sp1n1+1n2t = \frac{M_1 - M_2}{S_p \cdot \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}

Substitute the values:

t=56.652.613.57112+118t = \frac{56.6 - 52.6}{13.57 \cdot \sqrt{\frac{1}{12} + \frac{1}{18}}} t=413.570.0833+0.0556t = \frac{4}{13.57 \cdot \sqrt{0.0833 + 0.0556}} t=413.570.1389=413.570.372645.0550.791t = \frac{4}{13.57 \cdot \sqrt{0.1389}} = \frac{4}{13.57 \cdot 0.3726} \approx \frac{4}{5.055} \approx 0.791

Step 5: Degrees of Freedom (df)

To be conservative, we use the smaller sample size - 1 to estimate the degrees of freedom (df):

df=n11=121=11df = n_1 - 1 = 12 - 1 = 11

Step 6: p-value

Using a t-distribution table or software, look up the p-value for t=0.791t = 0.791 with 11 degrees of freedom in a one-tailed test at α=0.01\alpha = 0.01.

For t=0.791t = 0.791 and df=11df = 11, the corresponding p-value is approximately 0.2197.

Conclusion:

Since the p-value of 0.2197 is greater than the significance level α=0.01\alpha = 0.01, we fail to reject the null hypothesis. There is not enough evidence to support that μ1>μ2\mu_1 > \mu_2.

p-value0.2197\text{p-value} \approx 0.2197

Let me know if you want more details or have questions!


Here are 5 related questions:

  1. How do you interpret a p-value in the context of hypothesis testing?
  2. What assumptions are necessary for conducting a t-test?
  3. What is the difference between one-tailed and two-tailed tests?
  4. How does sample size affect the test statistic and p-value?
  5. What is the impact of unequal variances on this test?

Tip: Always double-check if your hypothesis test should be one-tailed or two-tailed, as this affects how the p-value is interpreted.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Two-Sample t-Test
Pooled Standard Deviation
Degrees of Freedom

Formulas

Pooled Standard Deviation: Sp = sqrt(((n1 - 1) * SD1^2 + (n2 - 1) * SD2^2) / (n1 + n2 - 2))
Test Statistic: t = (M1 - M2) / (Sp * sqrt(1/n1 + 1/n2))

Theorems

Central Limit Theorem
t-Distribution

Suitable Grade Level

Undergraduate Statistics