Surveys in a mid-sized Western town assessed the number of households with broad

Question

Surveys in a mid-sized Western town assessed the number of households with broadband Internet access. Results are as follows, with the number in each survey - 2250

Survey Date

2001

2004

2007

2008

Percentage of Homes with access

5%

24%

48%

55%

a)Display the data in a 2-way table of counts.

b)Test the null hypothesis that the proportion of homes that access the Internet using broadband has not changed over this period of time. Report the degrees of freedom, test statistic, and associated p-value.

Survey Date

2001

2004

2007

2008

Percentage of Homes with access

5%

24%

48%

55%

Explanation / Answer

1) for a 2 - way table

we need to categorize data in two segements. But here we have only one independent variable and one dependent variable.

Hence 2-way table of counts cannot be created

2) Null hypothesis : There is no increase in percentage over years

Alternate hypotheis: There is increase in percentage over years

Let us perform CHi-square test

CHi-square = SUM ((O-E)^2/E)

O - observed value

E - expected value

From the following table

Chi -square = 1073

As pwe significance level = 0.5

and degrees of freedom = n- 1 = 4-1 =3,

from chi-square table, chi value = 7.815

Since observed chi-square <<< calculated Chi - square

we can say that Null hypothesis is rejected

i.e. claim that there is no increase in percentage is not true

Survey % total no(2250*%/100)=observed expected (O-E)^2/E 2001 5 112.5 742.5 534.5455 2004 24 540 742.5 55.22727 2007 48 1080 742.5 153.4091 2008 55 1237.5 742.5 330 avg=742.5 sum=1073

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