Question

an urban economist is curious if the distribution in where oregon residents live is different today than it was in 1990. she observes that today there are approximately 3,109 thousand residents in nw oregon, 902 thousand residents in sw oregon, 244 thousand in central oregon, and 102 thousand in eastern oregon. she knows that in 1990 the breakdown was as follows: 72.7% nw oregon, 20.7% sw oregon, 4.8% central oregon, and 2.8% eastern oregon. can she conclude that the distribution in residence is different today at a 0.05 level of significance? enter the p - value - round to 4 decimal places. make sure you put a 0 in front of the decimal. p - value = __

Explanation:

Step1: Calculate the total number of residents today

$3109 + 902+244 + 102=4357$ (in thousands)

Step2: Calculate the expected number of residents in each region based on 1990 distribution

• NW Oregon: $E_{NW}=0.727\times4357 = 3177.539$
• SW Oregon: $E_{SW}=0.207\times4357 = 901.899$
• Central Oregon: $E_{C}=0.048\times4357 = 209.136$
• Eastern Oregon: $E_{E}=0.028\times4357 = 121.996$

Step3: Calculate the chi - square test statistic $\chi^{2}=\sum\frac{(O - E)^{2}}{E}$

• For NW Oregon: $\frac{(3109 - 3177.539)^{2}}{3177.539}=\frac{(- 68.539)^{2}}{3177.539}=\frac{4697.644}{3177.539}\approx1.48$
• For SW Oregon: $\frac{(902 - 901.899)^{2}}{901.899}=\frac{(0.101)^{2}}{901.899}=\frac{0.010201}{901.899}\approx0.000011$
• For Central Oregon: $\frac{(244 - 209.136)^{2}}{209.136}=\frac{(34.864)^{2}}{209.136}=\frac{1215.407}{209.136}\approx5.81$
• For Eastern Oregon: $\frac{(102 - 121.996)^{2}}{121.996}=\frac{(-19.996)^{2}}{121.996}=\frac{399.84}{121.996}\approx3.28$

$\chi^{2}=1.48+0.000011 + 5.81+3.28=10.570011$

Step4: Determine the degrees of freedom

$df=k - 1$, where $k = 4$ (number of regions), so $df=4 - 1=3$

Step5: Find the p - value

Using a chi - square distribution table or calculator with $\chi^{2}=10.570011$ and $df = 3$, the p - value is $P(\chi^{2}_{3}>10.570011)\approx0.0142$

Answer:

$0.0142$