Question

Car manufacturers are interested in whether there is a relationship between the size of car an individual drives and the number of people in the driver's family (that is, whether car size and family size are independent). To test this, suppose that 798 car owners were randomly surveyed with the following results. Conduct a test for independence at the 5% level.

Family Size	Sub & Compact	Mid-size	Full-size	Van & Truck
1	20	35	41	34
2	21	50	69	81
3 - 4	21	49	99	90
5+	19	29	70	70

• Part (a)

  State the null hypothesis.

  The size of the car an individual drives is dependent on the number of people in the driver's family.The size of the car an individual drives is independent of the number of people in the driver's family.    

• Part (b)

  State the alternative hypothesis.

  The size of the car an individual drives is dependent on the number of people in the driver's family.The size of the car an individual drives is independent of the number of people in the driver's family.    

• Part (c)

  What are the degrees of freedom? (Enter an exact number as an integer, fraction, or decimal.)

• Part (d)

  State the distribution to use for the test.

  χ²₃

  t₃

      

  χ²₉

  t₉

• Part (e)

  What is the test statistic? (Round your answer to two decimal places.)

• Part (f)

  What is the p-value? (Round your answer to four decimal places.)
  
  
  Explain what the p-value means for this problem.If

  H₀

  is false, then there is a chance equal to the p-value that the value of the test statistic will be equal to or less than the calculated value.If

  H₀

  is false, then there is a chance equal to the p-value that the value of the test statistic will be equal to or greater than the calculated value.    If

  H₀

  is true, then there is a chance equal to the p-value that the value of the test statistic will be equal to or greater than the calculated value.If

  H₀

  is true, then there is a chance equal to the p-value that the value of the test statistic will be equal to or less than the calculated value.

• Part (g)

  Sketch a picture of this situation. Label and scale the horizontal axis, and shade the region(s) corresponding to the p-value.

• Part (h)

  Indicate the correct decision ("reject" or "do not reject" the null hypothesis) and write the appropriate conclusion.(i) Alpha (Enter an exact number as an integer, fraction, or decimal.)
  α =
  
  (ii) Decision:

  reject the null hypothesisdo not reject the null hypothesis    

  
  (iii) Reason for decision:

  Since α > p-value, we reject the null hypothesis.Since α > p-value, we do not reject the null hypothesis.    Since α < p-value, we do not reject the null hypothesis.Since α < p-value, we reject the null hypothesis.

  
  (iv) Conclusion:

  There is sufficient evidence to conclude that the size of the car an individual drives is dependent on the number of people in the driver's family.There is not sufficient evidence to conclude that the size of the car an individual drives is dependent on the number of people in the driver's family.   

Answer 1

a) As we are testing here whether  there is a relationship between the size of car an individual drives and the number of people in the driver's family, therefore the null and the alternative hypothesis here are given as:

a) H0: There is no relationship between the size of car an individual drives and the number of people in the driver's family that is The size of the car an individual drives is independent of the number of people in the driver's family.   

b) Ha: The size of the car an individual drives is dependent on the number of people in the driver's family

c) The degrees of freedom here is computed as:

Df = (num of column - 1) (num of rows - 1) = (4 - 1)(4 - 1) = 9
Therefore 9 is the degrees of freedom here.

d) As 9 is degrees of freedom and we are doing a test of independence which is a chi square test, therefore the distribution here is given as:

$\chi^2_9$

e) For each of the 4*4 = 16 cells we first compute the expected value as:
E_i = (Sum of row i)(Sum of column i) / Grand Total

Post this, we compute the chi square test statistic contribution as:

$\chi^2_i = \frac{(O_i - E_i)^2}{E_i}$

These computations are made as follows:

1 2 Size Sub 20 (13.20) [3.51] 21 (22.43) [0.09] 21 (26.29) [1.06] 19 (19.08) [0.00] Compact Mid-size 35 (26.55) [2.69] 50 (45.14) [0.52] 49 (52.90) [0.29] 29 (38.40) [2.30] Full size 41 (45.45) [0.44] 69 (77.27) [0.88] 99 (90.55) [0.79] 70 (65.73) [0.28] Van and Truck 34 (44.80) [2.60] 81 (76.16) [0.31] 90 (89.25) [0.01] 70 (64.79) [0.42] 3-4 5+

The circular bracket values contain the expected values while the square bracket contains the chi square test statistic for that cell.

Now the chi square test statistic here is computed as:

$\chi^2 = \sum \chi^2_i = \sum \frac{(O_i - E_i)^2}{E_i} = 16.19$

Therefore 16.19 is the required chi square test statistic here.

f) The p-value for this one tailed test, is computed from the chi square distribution tables as:

$p = P(\chi^2_9 > 16.19) = 0.0631$

Therefore 0.0631 is the required p-value here.

The interpretation of the p-value is that given H0 is true, there is a probability equal to p-value to reject the null hypothesis that is the test statistic being greater than the critical value.

g) As we are doing the test at the 5% level of significance here, therefore we have here:

g= 0,05

This is the graph that represents the p-value here:

As the p-value here is 0.0631 > 0.05 which is the level of significance, therefore it lies in the rejection region here and we can Reject H0 here.

0.125 0.100 0.075 0.050 0.025 0.000 0 5 10 15 20 25

Therefore There is not sufficient evidence to conclude that the size of the car an individual drives is dependent on the number of people in the driver's family.