Answer:
- The value of the 2 test statistic is 6.98
- There is no sufficient evidence to reject the claim of the specific distribution
Step-by-step explanation:
Given
Proportion (p)
[tex]ABC = 29\% = 0.29\\CBS = 28\% = 0.28\\NBC = 25\% = 0.25\\Others = 18\% = 0.18\\[/tex]
Samples
[tex]ABC = 95\\CBD = 65\\NBC = 87\\Others = 53\\\alpha = 0.05[/tex]
First, we formulate a table for chi square totals:
[tex]x^2 = \frac{(O - E)^2}{E}[/tex]
Where
O = Observed Frequency and E = Expected Frequency
So, we have:
[tex]\begin{array}{cccccc}{\ } & {O} & {E = np} & {O - E} & {(O -E)^2} & {x^2 = \frac{(O - E)^2}{E}} \ \\ \\ {0.29} & {95} & {87} & {8} & {64} & {0.736} &{0.28} & {65} & {84} & {-19} & {361} & {4.300} & {0.25} & {87} & {75} & {12} & {144} & {1.920} & {0.18} & {53} & {54} & {-1} & {1} & {0.019}\ \end{array}[/tex]
For ABC
[tex]E = np = 300 * 0.29= 87[/tex]
[tex]O-E =95 - 87 = 8[/tex]
[tex](O-E)^2 = 8^2 = 64[/tex]
[tex]x^2 = \frac{(O - E)^2}{E} = \frac{64}{87} = 0.736[/tex]
For CBS
[tex]E =np = 300 * 0.28 = 84[/tex]
[tex]O -E = 65 - 84 = -19[/tex]
[tex](O -E)^2 = (-19)^2 = 361[/tex]
[tex]x^2 = \frac{(O - E)^2}{E} = \frac{361}{84} = 4.300[/tex]
For NBC
[tex]E =np = 300 * 0.25 = 75[/tex]
[tex]O -E = 87 - 75 = 12[/tex]
[tex](O -E)^2 = (12)^2 = 144[/tex]
[tex]x^2 = \frac{(O - E)^2}{E} = \frac{144}{75} = 1.920[/tex]
For others:
[tex]E = np = 300 * 0.18 =54[/tex]
[tex]O - E = 53 - 54 = -1[/tex]
[tex](O - E)^2 = (-1)^2= 1[/tex]
[tex]x^2 = \frac{(O - E)^2}{E} = \frac{1}{54} = 0.019[/tex]
[tex]\sum x^2 = 0.736 + 4.300 + 1.920 + 0.019[/tex]
[tex]\sum x^2 = 6.975[/tex]
[tex]\sum x^2 = 6.98[/tex] --- approximated
The value of the 2 test statistic is 6.98
Next, is to determine the p value:
Calculate the degree of freedom
[tex]df = n - 1[/tex]
[tex]df = 4 - 1[/tex]
[tex]df = 3[/tex]
Using the [tex]x^2[/tex] table in the 3rd row,
[tex]\sum x^2 = 6.98[/tex] corresponds to:
[tex]p = 0.072699[/tex]
The value of p is:
[tex]0.05 < P < 0.10[/tex]
This implies that, we fail to reject H o
