Question

can someone help me finish this? on excel please. forgot the steps

uestion 1. Problem 1 Contrib to X42 Hotline Callers (Oi) U.S. Population 19 22.9 35.8 22.5 Northeast Midwest South West 80 46 Total XA2 goes here 200 sum 100 Hyp Test, XA2:| e geographical Hyp Test, p-Value: continued: HO: The geographical distribution of the hotline distribution of the hotline callers is the same as the U.S. population. V. Concl. Since the test stat of p-Value ? falls above? below? the crit bound of V. Concl Since the test stat of XA2- falls above? below? the crit bound of callers is the same as the U.S. population. H1: The geographical distribution of the hotline distribution of the hotline chi-soldo not the U.S. population. alpha-0.05, do/do not reject HO w/ at least least 95% confidence 95% confidence. the U.S. population. Il. Rej. Reg. alpha = 0.05 Rej HO if p-Value <0.05 II. Rej. Reg. alpha 0.05 V. Implication V.Implication There is/is not enough There is/is not enough evidence to conclude the evidence to conclude the distrib of hotline callers distrib of hotline calers is different from the distrib is different from the distrib of the pop across the country Rej HO if xA2? Ill. Test Stat. of the pop across the country Il. Test Stat. Total XA2 goes here-CHIDIST(?.?) Use Chitest to check the test results here: CHITEST(C8:C11,D8:D11)

Answer 1

Solution:

Here, we have to use chi square test for goodness of fit.

Null hypothesis: H₀: The geographical distribution of the hotline callers is the same as the U.S. population.

Alternative hypothesis: H_a: The geographical distribution of the hotline callers is not same as the U.S. population.

We are given level of significance = α = 0.05

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

We are given

N = 4

Degrees of freedom = df = N – 1 = 3

α = 0.05

Critical value = 7.814728

(by using Chi square table or excel)

Decision rule: Reject H₀ if χ² > 7.814728

Calculation tables for test statistic are given as below:

	U.S. Population	Hotline Caller (O)	E	Contribution to Chi square
Northeast	19	39	38	0.026315789
Midwest	22.9	55	45.8	1.848034934
South	35.6	60	71.2	1.761797753
West	22.5	46	45	0.022222222
Total	100	200	200	3.658370699

Chi square = ∑[(O – E)^2/E] = 3.658370699

P-value = 0.300795291

(By using Chi square table or excel)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that the geographical distribution of the hotline callers is the same as the U.S. population.

Excel formulas for this test is given as below:

	U.S. Population	Hotline Caller (O)	E	Contribution to Chi square
Northeast	19	39	=38	=(H2-I2)^2/I2
Midwest	22.9	55	45.8	=(H3-I3)^2/I3
South	35.6	60	71.2	=(H4-I4)^2/I4
West	22.5	46	45	=(H5-I5)^2/I5
Total	=SUM(G2:G5)	=SUM(H2:H5)	=SUM(I2:I5)	=SUM(J2:J5)
			P-value	=CHIDIST(J6,3)
			Critical Value	=CHIINV(0.05,3)