Question

X and Y are believed to have the same distribution, ~NORM(30,2.5). A sample is taken from each (n_x=30, n_y=35). What is the probability that the absolute difference between any Xx and Xy will exceed 0.5? You take sample Xx=30.22 and Xy= 30.85.  What is the probability that the difference will be at least as far from the mean difference as these two samples are?

Answer 1

Solution:-

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

Null hypothesis: u₁ - u₂ = 0.50
Alternative hypothesis: u₁ - u₂ 0.50

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 = 0.62202
DF = 63

t = [ (x₁ - x₂) - d ] / SE

t = -0.2090

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 hypothesised difference between 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 63 degrees of freedom is more extreme than - 0.2090; that is, less than -0.2090 or greater than 0.2090.

P-value = P(t < - 0.2090) + P(t > 0.2090)

Use the t-value calculator for finding p-values.

P-value = 0.4175 + 0.4175

P-value = 0.835

Interpret results. Since the P-value (0.835) is greater than the significance level (0.05), we cannot reject the null hypothesis.

The probability that the difference will be at least as far from the mean difference as these two samples are 0.835.