Consider a scenario where you have two samples. Sample 1 contains 50 observation
ID: 3259134 • Letter: C
Question
Consider a scenario where you have two samples. Sample 1 contains 50 observations, has the sample average value of 38 and the sample standard deviation of 2.5. Sample 2 contains 50 observations, has the sample average of 39 and the sample standard deviation of 2.0. Based on this information, please conduct a 90% significance test for the hypothesis that the population mean2 exceeds the population mean1 by exactly 1.5. Please keep in mind that we don’t know the population standard deviations but CAN assume that they are equal.
Question 4 options:
The test fails to reject the null that mean2-mean1=1.5
The test rejects the null that mean2-mean1=1.5
None of the above
The test fails to reject the null that mean2-mean1=1.5
The test rejects the null that mean2-mean1=1.5
None of the above
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: ?2 - ?1 = 1.5
Alternative hypothesis: ?2 - ?1 ? 1.5
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 0.10. 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 = 0.453
DF = 50 + 50 -2
D.F = 98
t = [ (x1 - x2) - d ] / SE
t = - 2.21
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 98 degrees of freedom is more extreme than - 2.21; that is, less than - 2.21 or greater than 2.21.
Thus, the P-value = 0.0294
Interpret results. Since the P-value (0.0294) is less than the significance level (0.10), we cannot accept the null hypothesis.
The test rejects the null that mean2-mean1=1.5