Question

The following data is serum iron measured in the blood of a group of adult women. Serum iron is measured both before and after treatment. Can you prove, at 95% confidence, that the treatment is effective in increasing blood iron? To get full credit, state your null and alternate hypothesis, show how you calculated your test statistic, show your critical region, and state your conclusion: is the treatment effective? YES or NO Also, note that this problem should be done using a dependent sample (paired data approach). It’s OK to use Excel to get the mean and the standard deviation.

	After treatment	Before treatment	Difference = After - Before
	42	35	7
Feel free to use Excel to	55	40	15
calculate the sample mean	45	32	13
and standard deviation.	48	46	2
	58	45	13
Assume data is normal.	60	48	12
	55	44	11
	50	40	10
	52	38	14
	50	45	5
		Mean =
		Standard dev =
		n =

Answer 1

This is a simple problem related to comparing two samples from the same subjects (paired samples)

We shall conduct the paired sample t test to analysis the problem .

From cacluations we obtain the follwoing

Difference Sample 1 Sample 2 Sample 1 Sample 2 35 15 32 13 13 45 12 52 38 50 45 51.5 5.662 10 41.3 10.2 Average 4.237 St. Dev. 5.187 From the sample data, it is found that the corresponding sample means are X151.5 X2 41.3

Also, the provided sample standard deviations are: 1-5.662 2-5.187 and the sample size is n 10. For the score differences we have D 10.2 SD4.237 The following null and alternative hypotheses need to be tested: This corresponds to a right-tailed test, for which a t-test for two paired samples be used. (2) Reiection Recion Based on the information provided, the significance level is α-0.05, and the degrees of freedom are df 9. Hence, it is found that the critical value for this right-tailed test ist 1.833, for a0.05 and df-9. The rejection region for this right-tailed test is R t:t > 1.833.

The t-statistic is computed as shown in the following formula: t =--=--= 7.612 SD/Vn 4.237/V10 4 Since it is observed thatt 7.612t 1.833, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0, and since p = 0 < 0.05, it is concluded that the null hypothesis is rejected. 5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ| is greater than μ2, at the 0.05 significance level. Confidence Interval The 95% confidence interval is 7.169 < < 13.231.

p-value Test 1.6426820774518E-5 0.40 0.35 0.30 0.25 0.20 0.15- 0.10 .05 0.00 -4 -2

Thus we conclude that the after treatment impact is greater significantly and hence can say that treatment has been effective significantly at 95 % confidence level.