Question

you are conducting a goodness of fit test for the claim that a variable has the following distribution:

$h_{o}: p_{a}=0.15 ; p_{b}=0.3 ; p_{c}=0.4 ; p_{d}=0.15$

complete the table. report all answers accurate to three decimal places.

what is the chi - square test - statistic for this data?

$x^{2}=$

Explanation:

Step1: Calculate total observed frequency

Total observed frequency \(n = 12 + 19+41 + 21=93\)

Step2: Calculate expected frequency for each category

• For category A: \(E_A=n\times p_A=93\times0.15 = 13.95\)
• For category B: \(E_B=n\times p_B=93\times0.3 = 27.9\)
• For category C: \(E_C=n\times p_C=93\times0.4 = 37.2\)
• For category D: \(E_D=n\times p_D=93\times0.15 = 13.95\)

Step3: Calculate \(\frac{(O - E)^2}{E}\) for each category

• For category A: \(\frac{(12 - 13.95)^2}{13.95}=\frac{(- 1.95)^2}{13.95}=\frac{3.8025}{13.95}\approx0.273\)
• For category B: \(\frac{(19 - 27.9)^2}{27.9}=\frac{(-8.9)^2}{27.9}=\frac{79.21}{27.9}\approx2.839\)
• For category C: \(\frac{(41 - 37.2)^2}{37.2}=\frac{(3.8)^2}{37.2}=\frac{14.44}{37.2}\approx0.388\)
• For category D: \(\frac{(21 - 13.95)^2}{13.95}=\frac{(7.05)^2}{13.95}=\frac{49.7025}{13.95}\approx3.563\)

Step4: Calculate chi - square test - statistic

\(\chi^{2}=\sum\frac{(O - E)^2}{E}=0.273 + 2.839+0.388 + 3.563=7.063\)

Answer:

Category	Observed Frequency	Expected Frequency	\(\frac{(O - E)^2}{E}\)
B	19	27.9	2.839
C	41	37.2	0.388
D	21	13.95	3.563

\(\chi^{2}=7.063\)