Question

The null and alternate hypotheses are:

H₀ : μ₁ = μ₂
H₁ : μ₁ ≠ μ₂

A random sample of 12 observations from one population revealed a sample mean of 23 and a sample standard deviation of 2.5. A random sample of 5 observations from another population revealed a sample mean of 25 and a sample standard deviation of 2.7.

At the 0.10 significance level, is there a difference between the population means?

State the decision rule. (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.)

Compute the pooled estimate of the population variance. (Round your answer to 3 decimal places.)

Compute the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)

State your decision about the null hypothesis.

Do not reject H₀.

Reject H₀.

The  p-value is

between 0.2 and 0.1

less than 0.001

between 0.02 and 0.05

between 0.001 and 0.01

between 0.05 and 0.1

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

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

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.10. 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 = 1.4067
DF = 15
t = [ (x₁ - x₂) - d ] / SE

t = - 1.42

t_critical = + 1.753

Rejection region is - 1.753 > t > 1.753

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.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 15 degrees of freedom is more extreme than -1.42; that is, less than -1.42 or greater than 1.42.

Thus, the P-value = 0.176

The p-value is between 0.2 and 0.1.

Interpret results. Since the P-value (0.176) is greater than the significance level (0.10), we have to accept the null hypothesis.