Question

In a study of the development of the thymus gland, researchers weighted the gland of 10 chicken embryos. Five of the embryos had been incubated for 14 days, and five had been incubated for 15 days. The thymus weights were as shown in the table below.

14 days [29.6, 21.5, 28.0, 34.6, 44.9]

15 days [32.7, 40.3, 23.7, 25.2, 24.2]

The thymus weights of embryos are assumed to be normally distributed. Let μ1 be the mean thymus weights of embryos incubated for 14 days, and μ2 be the mean thymus weights of embryos incubated for 15 days.

a. Compute the sample mean and the sample variance for each group.

b. Assuming equal population variances, construct a 95% two-sided confidence interval for the difference of the two population means μ1 − μ2.

c.Assuming equal population variances, consider the hypothesis testing problem H0 : μ1 − μ2 = 0 againstHa : μ1 − μ2 ̸= 0. Use t-test and α = 0.05. Include details such as test statistic, rejection region, and decision of whether to reject H0.

d.Suppose someone questions about the assumption of equal population variances. Use t-test not assuming equalvariancestotestH0 :μ1−μ2 =0againstHa :μ1−μ2 ̸=0atα=0.05.

Answer 1

a) = (29.6 + 21.5 + 28 + 34.6 + 44.9)/5 = 31.72

s1^2 = ((29.6 - 31.72)^2 + (21.5 - 31.72)^2 + (28 - 31.72)^2 + (34.6 - 31.72)^2 + (44.9 - 31.72)^2)/4 = 76.197

= (32.7 + 40.3 + 23.7 + 25.2 + 24.2)/5 = 29.22

s2^2 = ((32.7 - 29.22)^2 + (40.3 - 29.22)^2 + (23.7 - 29.22)^2 + (25.2 - 29.22)^2 + (24.2 - 29.22)^2)/4 = 51.677

b) The pooled variance(s_p²) = ((n1 - 1)s1^2 + (n2 - 1)s2^2)/(n1 + n2 - 2)

                                            = (4 * 76.197 + 4 * 51.677)/(5 + 5 - 2)

                                            = 63.937

df = 5 + 5 - 2 = 8

At 95% confidence interval the critical value is t^* = 2.306

The 95% confidence interval for is

() +/- t^* * sqrt(s_p²/n1 + s_p²/n2)

= (31.72 - 29.22) +/- 2.306 * sqrt(63.937/5 + 63.937/5)

= 2.5 +/- 11.662

= -9.162, 14.162

c) The test statistic t = ()/sqrt(s_p²/n1 + s_p²/n2)

                                = (31.72 - 29.22)/sqrt(63.937/5 + 63.937/5)

                                = 0.494

At alpha = 0.05, the critical values are t_0.025,8 = +/- 2.306

Since the test statistic value is not greater than the positive critical value(0.494 < 2.306), so we should not reject H₀.

d) The test statistic t = ()/sqrt(s1^2/n1 + s2^2/n2)

                                 = (31.72 - 29.22)/sqrt(76.197/5 + 51.677/5)

                                 = 0.494

df = (s1^2/n1 + s2^2/n2)^2/((s1^2/n1)^2/(n1 - 1) + (s2^2/n2)^2/(n2 - 1))

   = (76.197/5 + 51.677/5)^2/((76.197/5)^2/4 + (51.677/5)^2/4)

   = 8

At alpha = 0.05, the critical values are t_0.025,8 = +/- 2.306

Since the test statistic value is not greater than the positive critical value(0.494 < 2.306), so we should not reject H₀.