Question

A study of iron deficiency among infants compared samples of infants following different feeding regimens. One group contained breast-fed infants, while the infants in another group were fed a standard baby formula without any iron supplements. Here are summary results on blood hemoglobin levels at 12 months of age. Group n x s Breast-fed 22 13.3 1.6 Formula 20 12.7 1.7 (a) Is there significant evidence that the mean hemoglobin level is higher among breast-fed babies? State H0 and Ha. H0: ?breast-fed ? ?formula; Ha: ?breast-fed < ?formula H0: ?breast-fed > ?formula; Ha: ?breast-fed = ?formula H0: ?breast-fed = ?formula; Ha: ?breast-fed > ?formula H0: ?breast-fed < ?formula; Ha: ?breast-fed = ?formula Carry out a t test. Give the P-value. (Use ? = 0.01. Use ?breast-fed ? ?formula. Round your value for t to three decimal places, and round your P-value to four decimal places.) t = P-value = What is your conclusion? Fail to reject the null hypothesis. There is significant evidence that the mean hemoglobin level is higher among breast-fed babies. Reject the null hypothesis. There is not significant evidence that the mean hemoglobin level is higher among breast-fed babies. Fail to reject the null hypothesis. There is not significant evidence that the mean hemoglobin level is higher among breast-fed babies. Reject the null hypothesis. There is significant evidence that the mean hemoglobin level is higher among breast-fed babies. (b) Give a 95% confidence interval for the mean difference in hemoglobin level between the two populations of infants. (Round your answers to three decimal places.) , (c) State the assumptions that your procedures in (a) and (b) require in order to be valid. We need sample sizes greater than 40. We need two independent SRSs from normal populations. We need the data to be from a skewed distribution. We need two dependent SRSs from normal populations.

Answer 1

Given:

	sample size (n)	Mean ( Given: a) ANSWER: C. H_0: ?breast-fed = ?formula H_a: ?breast-fed > ?formula b) Test statistic: t = X_1^bar - X_2^bar/squareroot S_1^2/n_1 + S_2^2/n_2 = 13.3 - 12.7/squareroot 1.6^2/22 + 1.7^2/20 = 1.175 Degrees of freedom: df = (S_1^2/n_1 + S_2^2/n_2)/(S_1^2/n_1)^2/n_1 - 1 + (S_2^2/n_2)^2/n_2 - 1 = (1.6^2/22 + 1.7^2/20)^2/(1.6^2/22)^2/22 - 1 + (1.7^2/20)^2/20 - 1 = 39 P-value: 0.1236 P-value > alpha, i.e 0.1236 > 0.01, That is Fail to Reject H_0 at 1% level of significance. ANSWER: C. Fail to reject the null hypothesis. There is not significant evidence that the mean hemoglobin level is higher among breast-fed babies. (b) Give a 95% confidence interval for the mean difference in hemoglobin level between the two populations of infants. (X_1^bar - Y_1^bar) plusminus t_df,alpha/2 * squareroot S_1^2/n_1 + S_2^2/n_2 (13.3 - 12.7) plusminus t_39,0.05/2 * squareroot 1.6^2/22 + 1.7^2/20 (-0.433,1.633) B. We need two independent SRSs from normal populations. )	SD (S)
Breast-fed	22	13.3	1.6
formula	20	12.7	1.7

a)

ANSWER: C.

H0: ?breast-fed = ?formula

Ha: ?breast-fed > ?formula

b) Test statistic:

$t = \frac{\bar{X1} - \bar{X2}}{\sqrt{\frac{S1^{2}}{n1} + \frac{S2^{2}}{n2}}} = \frac{13.3 - 12.7}{\sqrt{\frac{1.6^2}{22} + \frac{1.7^2}{20}}} = 1.175$

Degrees of freedom:

$df = \frac{\left ( \frac{S1^{2}}{n1} + \frac{S2^{2}}{n2} \right )^{2}}{\frac{\left ( \frac{S1^{2}}{n1} \right )^{2}}{n1-1} + \frac{\left ( \frac{S2^{2}}{n2} \right )^{2}}{n2-1}} = \frac{\left ( \frac{1.6^{2}}{22} + \frac{1.7^{2}}{20} \right )^{2}}{\frac{\left ( \frac{1.6^{2}}{22} \right )^{2}}{22-1} + \frac{\left ( \frac{1.7^{2}}{20} \right )^{2}}{20-1}} = 39$

P-value: 0.1236

P-value > $\alpha$ , i.e 0.1236 > 0.01, That is Fail to Reject Ho at 1% level of significance.

ANSWER: C. Fail to reject the null hypothesis. There is not significant evidence that the mean hemoglobin level is higher among breast-fed babies.

(b) Give a 95% confidence interval for the mean difference in hemoglobin level between the two populations of infants.

$(\bar{X1} - \bar{Y1}) \pm t_{df,\alpha /2} * \sqrt{\frac{S1^{2}}{n1} + \frac{S2^{2}}{n2}}$

$(13.3 - 12.7) \pm t_{39,0.05 /2} * \sqrt{\frac{1.6^{2}}{22} + \frac{1.7^{2}}{20}}$

(-0.433, 1.633)

B. We need two independent SRSs from normal populations.