Question

Dr. Mac, a radiologist, feels fairly certain that there are gender differences in whether or not people have ever had a broken bone. He surveyed 1,012 adults who were randomly selected from among all adults in the U.S. and he constructed the bivariate table below based on the survey results. Use the bivariate table to complete parts (a) -(h)

Ever had a broken bone?	Female	Male	Total
Yes	121	243	364
No	486	162	648
Total	607	405	1012

1. Calculate the degrees of freedom (3 points).

2. Determine the p-value for the chi-square statistic (5 points).

3. Decide whether there is evidence to reject the null hypothesis (use a= 0.01). Justify your decision (5 points).

  

4. Interpret the results of the hypothesis test in terms of the relationship between gender and broken bones (6 points).

Answer 1

Here

(a) Degree of freedom = (r - 1) * (c - 1) = (2 - 1) * (2 - 1) = 1

(b) Here first we have to find the expected value for each cell

Here

Expected value = Total Counts in the i^th row * Total ounts in the j^th column/ Total population

Expected value for Female & Yes = (607 * 364)/1012 = 218.33

Here similarly, we will calculate the expected values.

Ever had a broken bone?	Female	Male	Total
Yes	218.33	145.67	364
No	388.67	259.33	648
Total	607	405	1012

Now we will calculate chi-square statistic

X² =

Ever had a broken bone?	Female	Male	Total
Yes	43.39	65.03	108.42
No	24.37	36.53	60.90
Total	67.76	101.56	169.32

X² = 169.32

Here p - value would be calculated by using CHIDIST function

p - value = CHIDIST(X² < 169.32 , 1) = 0.0000

that is nearly equals to zero.

(c) There is enough evidence to reject the null hypothesis as p value is less than 0.01.

(d) Here we interpret the hypothesis as there is significant relationship between gender and broken bones.