Question

i need the last question.

Question Help To test the belief that sons are taller than their fathers, a student randomly selects 13 fathers who have adult male children. She records the height of both the father and son in inches and obtains the following data. Are sons taller than their fathers? Use the x = 0.05 level of significance. Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers Click here to view the table of data. Click here to view the table of critical t-values Which conditions must be met by the sample for this test? Select all that apply. YA. The sample size is no more than 5% of the population size. 3. The differences are normally distributed or the sample size is largo DC The sampling method results in an independent sample. D. The sampling method results in a dependent sample. GE. The sample size must be large. Let d -X-Y. Write the hypotheses for the test Ho: d=0 Hy: Hao Calculate the test statistic. to (Round to two decimal places as needed.)

Answer 1

The table given below ,

Height of Father (X)	Height of Son (Y)	di=X-Y	di^2
73.4	78.5	-5.1	26.01
67.8	71.3	-3.5	12.25
72.6	75	-2.4	5.76
71.6	73.4	-1.8	3.24
67.3	68.5	-1.2	1.44
72.4	73	-0.6	0.36
68.2	68.1	0.1	0.01
72.1	71.4	0.7	0.49
67.8	66.6	1.2	1.44
70.9	69	1.9	3.61
66.9	64.4	2.5	6.25
72.8	69.3	3.5	12.25
70.3	65.2	5.1	26.01
	Sum	0.4	99.12

From table ,

$\bar{d}=\frac{\sum d_i}{n}=\frac{0.4}{13}=0.03$

$s_d=\sqrt{\frac{\sum d_i^2-n*\bar{d}^2}{n-1}}=\sqrt{\frac{99.12-13*0.03^2}{13-1}}=2.8729$

Therefore , the value of the test statistic is,

$t_{o}=\frac{\bar{d}}{s_d/\sqrt{n}}=\frac{0.03}{2.8729/\sqrt{13}}=0.04$