A study was done using a treatment group and a placebo group. The results are sh

Question

A study was done using a treatment group and a placebo group. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.05 significance level for both parts.

A. Test the claim that the two samples are from populations with the same mean. What are the null and alternative hypotheses?

B. What is the test statistic, t?

C. What is the P Value?

D. Reject or fail to reject?

E. Construct a confidence interval suitable for testing the claim that the two samples are from populations with the same mean. (Round to 3 decimal places)

Explanation / Answer

Solution:-

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

Null hypothesis: 1 - 2 = 0,  the two samples are from populations with the same mean
Alternative hypothesis: 1 - 2 0,  the two samples are from populations with the different mean

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).

SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(0.552/25) + (0.872/35)] = 0.18364562147

DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (0.552/25 + 0.872/35)2 / { [ (0.552 / 25)2 / (24) ] + [ (0.872 / 35)2 / (34) ] }
DF = 0.0011374238 / (0.0000061004 + 0.00001375504) = 57.2852477709 or 57

t = [ (x1 - x2) - d ] / SE = [ (2.33 - 2.61) - 0 ] / 0.18364562147 = -1.5246756 or -1.525

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 57 degrees of freedom is more extreme than -1.5246756; that is, less than -1.5246756 or greater than 1.5246756.

We use the t Distribution Calculator to find P(t < -1.5246756)

The P-Value is 0.132789.
The result is not significant at p < 0.05.

Interpret results. Since the P-value (0.132789) is more than the significance level (0.05), we cannot reject the null hypothesis.

Conclusion. Fail to reject the null hypothesis. If have significant evidence to prove the claim that the two samples are from populations with the same mean.

Calculation for confidence interval :-

Pooled Variance
s2p = (SS1 + SS2) / (df1 + df2) = 1.06 / 58 = 0.02

Standard Error
s(M1 - M2) = ((s2p/n1) + (s2p/n2)) = ((0.02/25) + (0.02/35)) = 0.04

Confidence Interval
1 - 2 = (M1 - M2) ± ts(M1 - M2) = 0.28 ± (2 * 0.04) = 0.28 ± 0.0708

95% CI [0.2092, 0.3508]

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