Question

Let X be the weights of newborn babies in pounds, and denote m to be its median. For a sample of 14, we have: 5.2 11.8 7.1 14.4 6.3 7.5 11.1 5.5 9.8 12.7 10.5 10.9 8.7 9.2 (a) Give point estimates of 70.25, 0.35, M, 0.75 (b) Find the following confidence intervals and, from Table II in Appendix B, state the associated confidence coefficient: i. (X1), X(5)), a confidence interval for 0.25. ii. (X(3), X (8)), a confidence interval for 0.35 iii. (X(5), X (10)), a confidence interval for 10.5. iv. (X (8), X (13)), a confidence interval for 70.75 (c) Now, we would like to test H: m = 8 against H:m<8 at significance level a = 0.05. i. Use the sign test to test the hypothesis. ii. Use the Wilcoxon sign-rank test to test the hypothesis. ii. Use the t-test to test the hypothesis, assuming the underlying distribution is symmetric.

Answer 1

5.2
11.8
7.1
14.4
6.3
7.5
11.1
5.5
9.8
12.7
10.5
10.9
8.7
9.2

A -

Step 1. Arrange the data in ascending order: 5.2, 5.5, 6.3, 7.1, 7.5, 8.7, 9.2, 9.8, 10.5, 10.9, 11.1, 11.8, 12.7, 14.4

Step 2. Compute the position of the p^th percentile (index i):

i = (p / 100) * n), where p = 25 and n = 14

i = (25 / 100) * 14 = 3.5

Step 3. The index i is not an integer, round up. (i = 4) ⇒ the 25^th percentile is the value in 4^th position, or 7.1

Answer: the 25^th percentile is 7.1

25^th percentile = 7.1

Similarly,

35^th percentile = 7.5

50^th percentile = 9.5

75^th percentile = 11.1

C -

1.

Values more than 8 = 9

Values less than 8 = 5

The test statistic X is min(5,9) = 5

The critical value for the significance level provided and the type of tail is X* = 3

Since in this case X = 5 & gt; X* = 3, X=5>X*=3, there is not enough evidence to claim that the population median of differences is less than 8, at the 0.05 significance level.

2.

Dividing the sample in two parts - the left and right seven observations (assuming symmetric distribution)

The sum of positive ranks is:

W^+ = 1+2+3+4+5+6+7 = 28

and the sum of negative ranks is:

W^- = 0

Hence, the test statistic is T = min{W^+, W^-\} = 0

The critical value for the significance level alpha = 0.05, and the type of tail specified is T^* = 3, and the null hypothesis is rejected if T≤3

Since in this case T = 0 ≤ 3, there is enough evidence to claim that the population median of differences is less than 8, at the alpha = 0.05 significance level

(Assuming symmetric distribution for two groups)

3.

Assuming symmetric distribution for two groups -

The U-value is 0.

The critical value of U at p < .05 is 11.

Therefore, the result is significant at p < .05.

2.

For CI of a median,

The lower 95% CI is given by the n/2 - 1.96 * sqrt(n) / 2 th ranked value

The upper 95% CI is given by the 1 + n/2 + 1.96 * sqrt(n) / 2 th ranked value

For 25th percentile -

lower 95% CI = 4/2 - 1.96 * sqrt(4) / 2 th ranked value = 1st ranked value

upper 95% CI = 1 + 4/2 + 1.96 * sqrt(4) / 2 th ranked value = 5th ranked value

Similarly for 35th percentile - 3rd and 8th value

Similarly for 50th percentile - 5th and 10th value

Similarly for 75th percentile - 8th and 13th value