There are two teaching methods. Method A and Method B. Five students were taught

Question

There are two teaching methods. Method A and Method B. Five students were taught with Method A and f > other students with method B. The sample mean for A is 75, and the sample standard deviation is 5, the sample mean for B is 80 and the sample standard deviation is 5. a. Test the null hypothesis H_0: mu_1 = mu_2 vs, the alternative H_1 mu_1 notequalto mu_2. Use alpha = 1 b. Construct a 90 percent confidence interval for the difference between the means c. What assumptions are needed about the two samples?

Explanation / Answer

Solution:-

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

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

Null hypothesis: 1 = 2
Alternative hypothesis: 1 2

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 = sqrt[(52/5) + (52/5)] = 3.16227766

DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (52/5 + 52/5)2 / { [ (52 / 5)2 / (4) ] + [ (52 / 5)2 / (4) ] }
DF = 100 / (6.25 + 6.25) = 8

t = [ (x1 - x2) - d ] / SE = [ (75 - 80) - 0 ] / 3.16227766 = -1.58

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 8 degrees of freedom is more extreme than -1.58; that is, less than -1.58 or greater than 1.58.

We use the t Distribution Calculator to find P(t < -1.58), and P(t > 1.58)

The P-Value is 0.152762.
The result is not significant at p < 0.10

Interpret results. Since the P-value (0.153) is greater than the significance level (0.10), we cannot reject the null hypothesis.

Conclusion. Fail to reject the null hypothesis. At given confidence interval there is no difference between the two means.

(b)

Calculation

Pooled Variance
s2p = (SS1 + SS2) / (df1 + df2) = 50 / 8 = 6.25

Standard Error
s(M1 - M2) = ((s2p/n1) + (s2p/n2)) = ((6.25/5) + (6.25/5)) = 1.58

Confidence Interval
1 - 2 = (M1 - M2) ± ts(M1 - M2) = 5 ± (1.86 * 1.58) = 5 ± 2.94

1 - 2 = (M1 - M2) = 5, 90% CI [2.06, 7.94].

You can be 90% confident that the difference between your two population means (1 - 2) lies between 2.06 and 7.94.

(c)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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