We have the age distribution in percentages for the whole United States accordin
ID: 3130522 • Letter: W
Question
We have the age distribution in percentages for the whole United States according to the 2010 US Census in the following table. We also have the age distribution in individuals for a simple random sample of 1000 Californians in 2010. At ? = 0.05, you will need to test if the Californian sample's age distribution is the same as the U.S. overall age distribution in 2010.
(a) Form the null and alternative hypotheses H? and H?.
(b) Specify appropriate equations for the ?² statistic and degrees of freedom df.
(c) Based on the U.S. overall age distribution, calculate expected frequencies.
(d) Given the expected and observed frequencies, calculate the ?² statistic and its df.
(e) Given ? = 0.05 and df, use CHISQ.INV(*,*) to look up for the right?tail ?²* critical value.
(f) Compare ?² statistic with ?²* critical to reject or accept H?. Make a verbal statement. (g)[1] Find the p?value of the test. Comment on the p?value
US% 6.54% 6.59% 6.70% 7.14% 6.99% 6.83% 6.47% 6.54% 677% 7.36% 722% 6.37% 5.45% 4.03% 3,01% 2.37% 1.86% 1.78% 100.00% CA Sample 68 67 70 76 74 74 69 69 70 72 69 59 49 35 26 21 16 16 1000 e Range Under 5 years 5 to 9 years 10 to 14 years 15 to 19 years 20 to 24 years 25 to 29 years 30 to 34 years 35 to 39 years 40 to 44 years 45 to 49 years 50 to 54 years 55 to 59 years 60 to 64 years 65 to 69 years 70 to 74 years 75 to 79 years 80 to 84 years 85 years and over TOTALExplanation / Answer
answer of part(a)
Ho: Californian sample's age distribution is the same as the U.S. overall age distribution in 2010
H1:Californian sample's age distribution is not the same as the U.S. overall age distribution in 2010
answer of part(b)
chi-squre statstic=sum((O-E)2/E) with df (n-1), wher n is number of group (here n=18) so df =17
answer of part(c)
in the above table expected frequency is column number 4
answer of part(d)
the chi-sqrue statistic is coulmn number 7 and row number 22 i.e 5.017 and df=17
answer of part (e) at alpha=0.05 to look right-tail chi-squre will be one-tail and chisq.inv(.1,17)=24.769
as for one tail the signicance level is double
answer of part(f) the two tail chi-squre critical value at alpha=0.05 and df=17 =27.587 as the alternative hypothesis is two tailed
we accept null hypothesis Ho as the critical value is more than calculated value
answer of part g)
p-value=0.9971 as p-value is more than alpha=0.05 so we accept null hypothesis
1 1 2 3 4 5 6 7 2 observed (O) Expected (E) 3 Range US% CA Sample Freq. O-E (O-E)2 (O-E)2/E 4 under 5 6.54 68 65.4 2.6 6.76 0.103364 5 5_9 6.59 67 65.9 1.1 1.21 0.018361 6 10_14 6.7 70 67 3 9 0.134328 7 15-19 7.14 76 71.4 4.6 21.16 0.296359 8 20-24 6.99 74 69.9 4.1 16.81 0.240486 9 25-29 6.83 74 68.3 5.7 32.49 0.475695 10 30-34 6.47 69 64.7 4.3 18.49 0.285781 11 35-39 6.54 69 65.4 3.6 12.96 0.198165 12 40-44 6.77 70 67.7 2.3 5.29 0.078139 13 45-49 7.36 72 73.6 -1.6 2.56 0.034783 14 50-54 7.22 69 72.2 -3.2 10.24 0.141828 15 55-59 6.37 59 63.7 -4.7 22.09 0.346782 16 60-64 5.45 49 54.5 -5.5 30.25 0.555046 17 65-69 4.03 35 40.3 -5.3 28.09 0.697022 18 70-74 3.01 26 30.1 -4.1 16.81 0.558472 19 75-79 2.37 21 23.7 -2.7 7.29 0.307595 20 80-84 1.86 16 18.6 -2.6 6.76 0.363441 21 above 85 1.78 16 17.8 -1.8 3.24 0.182022 22 Sum 100.02 1000 1000.2 -0.2 251.5 5.017669Related Questions
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