Question

The weight gain of women during pregnancy has an important effect on the birth weight of their children. If the weight gain is not adequate, the infant is more likely to be small and will tend to be less healthy. In a study conducted in two countries, weight gains (in kilogram) of women during the third trimester of pregnancy were measured. The results are summarized in the following table.

Country	n	ȳ	s2
Egypt	6	3.8	5.76
Kenya	10	3.3	3.61

     (a) Test for any difference on the average weight gains between two countries using α = .01. State the null and alternate hypotheses, and report the value of the test statistic, and the critical value used to conduct the test. Report your decision regarding the null hypothesis and your conclusion in the context of the problem. Consider two cases, equal and unequal variances.

  

     (b) Find a 95% confidence interval for µ1 − µ2 and interpret this interval. Assume population variances are equal.

    

Answer 1

a)Null hypothesis : Average weight gain of women during the third trimester of pregnancy in Egypt (mu 1) = Average weight gain of women during the third trimester of pregnancy in Kenya (mu 2)

mu 1 = mu 2

Or mu 1 - mu 2 = 0

Alternative hypothesis : mu 1 is not equal to mu 2

Or mu 1 - mu 2 is not equal to zero.

1) For equal variances, the test statistic is given by -

t = (sample mean weight gain of Egypt - sample mean weight gain of Kenya)/pooled standard deviation of both the sample * √[(1/sample size of first sample) + (1/sample size of second sample)]

= (Xbar - Ybar)/s*√[(1/nx) + (1/my)]

Here, pooled standard deviation (s)

= √[{(nx -1)*sx^2 + (ny-1)sy^2}/(nx + ny - 2)]

= √ [(5*5.76 + 9*3.61)/(6+10-2)]

= √ 61.29/14

= √4.378

= 2.092

Hence, the value of the test statistic will be -

t = (3.8-3.3)/2.092* √[ (1/6) +(1/10)]

= 0.5 /1.079

= 0.463

Degrees of the freedom = df = 6+10-2 = 14

Now, critical value of t for two tailed test with 14 degrees of freedom at 0.01 level of significance is 2.976

Since, the value of the calculated test statistic is less than the critical value of t, the null hypothesis may not be rejected suggesting that the mean weight gain of women in the third trimester of pregnancy are equal in both the countries.

2) For unequal Variance, test statistic is given by -

t = (Xbar - Ybar)√[(sx^2/nx) + (sy^2/ny)]

= 0.5/√[(5.76/6) + (3.61/10)]

= 0.5/√1.321

= 0.5/1.149

= 0.425

Here, also the value of the test statistic is less than the critical value of t with 14 degrees of freedom at 0.01 level of significance, hence, the null hypothesis may not be rejected, hence, there is no significant difference between the average weight gain of women in the third trimester of pregnancy in the two countries.

b) The 95% confidence interval for the difference between the two countries mean weight gain will be given by -

(X bar - Y bar) plus minus margin of errror

= [(3.8-3.3) plus minus t(0.05 level of significance, 14 df, two tailed)*s*√(1/nx) + (1/ny)]

The critical value of t with 14 df at 0.05 level of significance for the two tailed test at 0.05 level of significance is 2.144

= [0.5 plus minus 2.144 * 1.079]

=[0.5 plus minus 2.313]

=[0.5-2.313, 0.5 + 2.313]

= [-1.813, 2.813]

Hence, we are 95% confident that the difference between the average weight gain of women in the third trimester of pregnancy in both the countries lie between -1.813 and 2.813.

Also, since the interval include the null value that is zero, it suggests there is no significant difference between the two means.