Suppose two independent samples are associated with the following information: S

Question

Suppose two independent samples are associated with the following information: Sample 1: mean = 158, sample size = 22, SD = 2.06; Sample 2: mean = 157.4, sample size = 26, SD = 2.59. [From Sal kind e5, p. 214] (a) Using the formula on p. 202, compute the test statistic t (the obtained value) by hand. Keep lots of decimals until the end, and then round to two decimal places. How many degrees of freedom are there? (What is df?) Let's perform a two-tailed tailed test at the alpha = 0.05 level of significance. Find the critical value in Table B2. (If you cannot find an entry with the exact df, then choose the neighboring one that is closest to your actual total sample size, nl+n2.) Give the exact decimal places that you find in the table. Compare the obtained value to the critical value. What is your conclusion? We fail to reject the null hypothesis, because p 0.05. We reject the null hypothesis, because p > 0.05. We reject the null hypothesis, because p

Explanation / Answer

Solution:-

1 = 158, n1 = 22, S.D1 = 2.06

2 = 157.4, n2 = 26, S.D2 = 2.59

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

Null hypothesis: 1 - 2 = 0

Alternative hypothesis: 1 - 2 0

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).

S.E = sqrt[(s12/n1) + (s22/n2)]

S.E = 0.6715

D.F = 45.8475

D.F = 46

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

t = 0.8935

tcritical = 2.013

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 46 degrees of freedom is more extreme than - 0.89; that is, less than - 0.89 or greater than 0.89.

Thus, the P-value = 0.189 + 0.189 = 0.378

Interpret results. Since the P-value (0.378) is greater than the significance level (0.05), we have to accept the null hypothesis.

We fail to reject the null hypothesis beacuse p > 0.05

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.