Consider the following competing hypotheses and accompanying sample data. Use Ta

Question

Consider the following competing hypotheses and accompanying sample data. Use Table 2. H0: 1 – 2 = 6 HA: 1 – 2 6 12formula21.mml = 53 12formula22.mml = 31 s1 = 21.4 s2 = 15.1 n1 = 21 n2 = 17 Assume that the populations are normally distributed with unknown but equal variances. a. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) Test statistic b. Using the p-value approach, test the above hypotheses at the 1% significance level. The p-value is . H0. At the 1% significance level, we conclude that the difference between the means differs from 6. c. Repeat the analysis using the critical value approach. (Round your answer to 3 decimal places.) The critical value is . H0.

Explanation / Answer

Solution:-

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

Null hypothesis: 1 - 2 = 6
Alternative hypothesis: 1 -  2 6

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 0.01. Using sample data, we will conduct a two-sample z-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = sqrt[(s12/n1) + (s22/n2)]
SE = 5.935
z = [ (x1 - x2) - d ] / SE

z = 2.696

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 z statistic is more extreme than -2.696; that is, less than -2.696 or greater than 2.696..

Thus, the P-value = 0.014

Interpret results. Since the P-value (0.014) is greater than the significance level (0.01), we have to accept the null hypothesis.

zcritical = + 2.81

Rejectio region is z < - 2.81 or z > 2.81

Interpret results. Since the z-value (2.696) does not lies in the rejection region, hence we have to accept the null hypothesis.

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