Question

A random sample of 49 measurements from one population had a sample mean of 16, with sample standard deviation 3. An independent random sample of 64 measurements from a second population had a sample mean of 18, with sample standard deviation 4. Test the claim that the population means are different. Use level of significance 0.01. (a) What distribution does the sample test statistic follow? Explain.

The standard normal. We assume that both population distributions are approximately normal with known standard deviations. The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.     The Student's t. We assume that both population distributions are approximately normal with known standard deviations. The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.

(b) State the hypotheses.

H₀: μ₁ ≠ μ₂; H₁: μ₁ = μ₂H₀: μ₁ = μ₂; H₁: μ₁ ≠ μ₂     H₀: μ₁ = μ₂; H₁: μ₁ < μ₂H₀: μ₁ = μ₂; H₁: μ₁ > μ₂

(c) Compute

x₁ − x₂.

x₁ − x₂ =

Compute the corresponding sample distribution value. (Test the difference μ₁ − μ₂. Round your answer to three decimal places.)

Answer 1

A random sample of 49 measurements from one population had a sample mean of 16, with sample standard deviation 3. An independent random sample of 64 measurements from a second population had a sample mean of 18, with sample standard deviation 4. Test the claim that the population means are different. Use level of significance 0.01. (a) What distribution does the sample test statistic follow? Explain.

The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.   

(b) State the hypotheses.

H₀: μ₁ = μ₂; H₁: μ₁ ≠ μ₂    

(c) Compute

x₁ − x₂.

x₁ − x₂ =16-18= -2

Compute the corresponding sample distribution value. (Test the difference μ₁ − μ₂. Round your answer to three decimal places.)

T test value = -2.925

DF = n₁+n₂-2 =111

Table value of t with 111 DF at 0.01 level = 2.621

Rejection Region: Reject Ho if t < -2.621 or t > 2.621

Calculated t = -2.925,falls in the rejection region

The null hypothesis is rejected.

Pooled-Variance t Test for the Difference Between Two Means
(assumes equal population variances)
Data
Hypothesized Difference	0
Level of Significance	0.01
Population 1 Sample
Sample Size	49
Sample Mean	16
Sample Standard Deviation	3
Population 2 Sample
Sample Size	64
Sample Mean	18
Sample Standard Deviation	4
Intermediate Calculations
Population 1 Sample Degrees of Freedom	48
Population 2 Sample Degrees of Freedom	63
Total Degrees of Freedom	111
Pooled Variance	12.9730
Standard Error	0.6837
Difference in Sample Means	-2.0000
t Test Statistic	-2.9252
Two-Tail Test
Lower Critical Value	-2.6208
Upper Critical Value	2.6208
p-Value	0.0042
Reject the null hypothesis