sample of 40 observations is selected from one population with a population standard deviation of 4.6. The sample mean is 101.0. A sample of 47 observations is selected from a second population with a population standard deviation of 4.0. The sample mean is 99.3. Conduct the following test of hypothesis using the 0.10 significance level.
I don't know which test you are using. Try this one.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability
of the Z score.
To conduct the hypothesis test, we need to state the null hypothesis (H0) and the alternative hypothesis (Ha). In this case, we want to compare the means of two populations, so the hypotheses will be:
H0: μ1 - μ2 = 0
Ha: μ1 - μ2 ≠ 0
Where μ1 is the population mean of the first population and μ2 is the population mean of the second population.
We are given the sample means (x̄1 = 101.0, x̄2 = 99.3), the sample sizes (n1 = 40, n2 = 47), and the population standard deviations (σ1 = 4.6, σ2 = 4.0).
To conduct the test, we will use the two-sample t-test because the population standard deviations are known. The formula for the test statistic is:
t = (x̄1 - x̄2) / √((σ1^2/n1) + (σ2^2/n2))
Substituting the given values, we have:
t = (101.0 - 99.3) / √((4.6^2/40) + (4.0^2/47))
Calculating this expression, we find:
t ≈ 1.667
Next, we need to determine the critical value(s) for the t-distribution at a 0.10 significance level. Since we have a two-tailed test, we need to split the significance level in half (0.10 / 2 = 0.05). Using a t-table or calculator, we find that the critical value for a 0.05 significance level with degrees of freedom equal to the smaller sample size minus one (df = min(n1-1, n2-1)) is approximately ±1.987.
Since our test statistic (1.667) does not fall in the critical region (|t| > 1.987), we fail to reject the null hypothesis.
Thus, at the 0.10 significance level, there is insufficient evidence to conclude that the means of the two populations are significantly different.