Question

Assume that sample data, based on two independent samples of size n1=41 and n2=90, give us top enclose x subscript 1= 3.61, top enclose x subscript 2= 2.95, s1 = 0.19, and s2 = 0.22. Assume unequal population variances. Find 95% confidence interval for u2-u1. Conduct a test of hypotheses as an alternative to the method in the previous part. Provide the null and alternative hypotheses, test statistic, P-value and conclusion at 0.05 level.

Answer 1

Solution:-

a) 95% confidence interval for u2-u1 is C.I = (0.5846, 0.7345).

C.I = 0.66 + 1.979 × 0.03766

C.I = 0.66 + 0.07453

C.I = (0.5846, 0.7345)

b)

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

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

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

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 = 0.03766
DF = 129
t = [ (x₁ - x₂) - d ] / SE

t = 17.53

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 129 degrees of freedom is more extreme than -17.53; that is, less than -17.53 or greater than 17.53.

Thus, the P-value = less than 0.0001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we cannot accept the null hypothesis.