Question

Duehour, 20 minute Due The 736 2030 113 pm Questions Is there to be color and body type The below shows the results of a researcher's orbis of randomly selected people C Question 10/10) Question 2 (93/10) Question 3 (10) Question 4 (0/0) Question 5 (0/10) Question 60/10) C Question 7 (5/10) C Question (10) Question 9 [10/10 Question 101010 Frequencies of Hair Colors for Varis Body Types Blonde Brunette Red Head Short and leader R2 36 56 Short and Pede 70 72 Tolland Slender 55 107 76 Tillander 60 109 75 What can be concluded the a=0.10 significance level? Grade 43 1/100 What is the correct statistical test to use? Independence Paired t-test Homogenity Goodness-of-Fit 6. What are the mall and alternative hypotheses? Hair color and body type we dependent The distribution of hair color is the same for each body type Hair color and body type are independent The distribution of hair color is not the same for each body type The distribution of hair color is the same for each body type Hair color and body type are dependent Hair color and body type are independent The distribution of hair color is not the same for each body type c. The statistic for this data Please show your answer to three decimal places) & The p-value for dus sample Please show your newer to four decimal places) e. The p-value is Seladan Based on this, we should fail to reject the mill accept the will

Answer 1

a) As we are testing here whether the two variables are associated here, therefore this is a case of a test of independence here.

b) As we are testing here whether the two variables are associated here, therefore the null hypothesis here is given as:

H0: Hair colour and Body type are independent.

As we are testing here whether the two variables are associated here, therefore the alternative hypothesis here is given as:

Ha: Hair colour and Body type are dependent.

c) The expected frequency for each cell here is computed as:
E_i = (Sum of row i )*(Sum of column i) / Grand Total

Also the chi square test statistic contribution for each cell here is obtained as:

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

The computations are as shown here:

Short and Slender Short and Podgy Tall and Slender Tall and heavy Blonde 82 (62.24) [6.28] 70 (70.85) [0.01] 55 (66.12) (1.87] 60 (67.79) [0.90] Brunette Red Head 86 (96.73) [1.19] 56 (65.03) [1.25] 113 (110.12) [0.08] 72 (74.03) [0.06] 107 (102.78) [0.17] 76 (69.10) [0.69] 109 (105.37) [0.13] 75 (70.84) [0.24]

The circular bracket here contains the expected frequencies while the square bracket contains the chi square test statistic contributions here. Therefore the chi square test statistic here is computed as:

$\chi^2 = \sum \chi^2_i = \sum \frac{(O_i -E_i)^2}{E_i} = 12.863$

Therefore 12.863 is the chi square test statistic value here.

d) Df = (num of column - 1)*(Num of rows - 1) = (3 - 1)(4 - 1) = 6

Therefore the p-value here is obtained from chi square distribution tables as:

$p = P(\chi^2_6 > 12.863) = 0.0453$

Therefore 0.0453 is the required p-value here.

e) The p-value here is 0.0453 < 0.1, which is the level of significance here.

f) As the   The p-value here is 0.0453 < 0.1, therefore the test is significant here, therefore we Reject the null hypothesis H0 here.