Answer To: Week 14 Discussion Instructions Here we discuss the Chi-square goodness of fit and Chi-square test...
Pooja answered on May 09 2021
Week 14 Discussion Instructions
Here we discuss the Chi-square goodness of fit and Chi-square test of independence. This will be a detailed discussion worth 15 points so plan accordingly. Please use the following Instructions to write your reply.
Please follow the videos at
· Stats 3e Screencast 17.3
· Stats 3e Screencast 17.9
The Chapter 18 videos are optional. For your reply, please:
1. Describe a scenario where you could use the Chi-square goodness of fit test OR the Chi-square test of independence.
Suppose I want to test if gender is associated with a high salary group. A Chi-square test of Independence can be applied in this case. The two variables of concern are Gender and salary. Gender is categorized as male and female. The salary group is categorized as low level, average level, and high-level salary. The null hypothesis would be, that Gender and level of salary are independent of each other. And alternative hypothesis is that Gender and level of salary are dependent on each other.
2. Set up your test in SPSS and find out at what sample size the differences between groups become significant or at what counts per cell in the Chi-square test of independence the results become significant.
1. The point here is to get practice running one of the Chi-square methods and to see what relationship there is between sample size and the proportions you are comparing. For example, if we go into a typical online classroom of 20 people and ask people whether they prefer one of three pizza toppings (cheese only, vegetables only, meat only), we might propose to expect an even distribution of scores, so 33% per category.
Suppose I want to test if 4 categories of colors in M&M packet are equally likely. The null hypothesis is that all 4 colors are equally likely. V/s alternative hypothesis that at least one of the colors differs significantly. The scene 1 consists of a sample size is 101 with observed frequencies of 17 for Orange, 19 for Pink, 40 for yellow and 25 for red. The results are significant in scene 1 with Chi square (3) = 12.86, p=.004 (less than 5%). The scene 2 consists of sample size is 50 with observed frequencies of 10 for Orange, 15 for Pink, 10 for yellow and 15 for red. The results are not significant in scene 2 with Chi square (3) = 2, p=.57 (greater than 5%).
Reference calculations:
Scene 1: Statistically significant results.
ho: all 4 colors are equally likely. V/s h1: at least one of the colors differs significantly.
Oi
pi = 1/4
Ei = pi*N
(Oi-ei)^2/Ei
17
0.2500
25.25
2.70
19
0.2500
25.25
1.55
40
0.2500
25.25
8.62
25
0.2500
25.25
0.00
SUM
101
1
101
12.86
Test Statistic, chisq= 12.861 = sum(Oi-Ei)^2/Ei
Alpha= 0.05
k= 4.00
critical value = CHISQ.INV.RT(0.05,4-1) = 7.815
p-value = 0.004946159 = CHISQ.TEST(B2:B4,D2:D4)
Reject Ho as Chisq > critical value and p<5%. Conclude that all 4 colors are equally likely.
Scene 2: Statistically in-significant results.
ho: all colors are equally likely. V/s h1: at least one of the colors differs significantly
Oi
pi = 1/4
Ei = pi*N
(Oi-ei)^2/Ei
10
0.2500
12.5
0.50
15
0.2500
12.5
0.50
10
0.2500
12.5
0.50
15
0.2500
12.5
0.50
SUM
50
1
50
2.00
chisq= 2.000 = sum(Oi-Ei)^2/Ei
Alpha= 0.05
k = 4.00
critical value = CHISQ.INV.RT(0.05,4-1) = 7.815
p-value = 0.572406704 = CHISQ.TEST(B2:B4,D2:D4)
Fail to reject Ho as Chisq < critical value and p-value>5%. And conclude that at least one of the colors differs significantly
2. At 20 people, this is just over 6 people per group, so we might expect frequencies of Cheese (6 people), Vegetable (6 people), and Meat (8 people). We would then compare our actual distribution of people's responses to these expectations using SPSS or an online Chi-square goodness of fit calculator to see if the counts per cell (frequencies or proportions) do or don't match the expected distribution of responses. If our observed data is Cheese(n=3), Vegetable (n=3), and Meat (n=14), the Chi-square value is: The Chi^2 value is 7.5. The p-value is .02352. The result is significant at p < .05.
Suppose I want to test if gender and favorite color of M&M...