Question

Consider the following hypothesis test.

H₀: μ₁ − μ₂ = 0

H_a: μ₁ − μ₂ ≠ 0

The following results are from independent samples taken from two populations.

Sample 1	Sample 2
n1 = 35	n2 = 40
x1 = 13.6	x2 = 10.1
s1 = 5.9	s2 = 8.1

(a)

What is the value of the test statistic? (Use

x₁ − x₂.

Round your answer to three decimal places.)

(b)

What is the degrees of freedom for the t distribution? (Round your answer down to the nearest integer.)

(c)

What is the p-value? (Round your answer to four decimal places.)

p-value =  

Answer 1

Solution:-

Null hypothesis: u₁ - u ₂ = 0
Alternative hypothesis: u₁ - u ₂ 0  

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 1.6232

t = [ (x₁ - x₂) - d ] / SE

a)

t = 2.156

b)

DF = 35 + 40 - 2

D.F = 73

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

c)

Since we have a two-tailed test, the P-value is the probability that a t statistic having 73 degrees of freedom is more extreme than 2.156; that is, less than -2.156 or greater than 2.156.

P-value = P(t < - 2.156) + P(t > 2.156)

P-value = 0.0172 + 0.0172

Thus, the P-value = 0.0344

Interpret results. Since the P-value (0.0342) is less than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that there is a difference in the results from two populations.