Among the four northwestern states, Washington has 51% of the total population, Oregon has 30%, Idaho has 11%, and Montana has 8%. A market researcher selects a sample of 1000 subjects, with 450 in Washington, 340 in Oregon, 150 in Idaho, and 60 in Montana. At the 0.05 significance level, test the claim that the sample of 100 subjects has a distribution that agrees with the distribution of state populations.

Step-by-step explanation:

H₀ : The distribution of the sample agrees with the population distribution.

H₁ : the distribution of the does not agree with the population distribution.

Using the  Chi-square test statistics

we will compute the expected frequencies

Now expected frequency for Washington will be

Fe = Npi = 1000×51% = 5100

expected frequency for Oregon

Fe = Npi = 1000×30% = 300

expected frequency for Idaho

Fe = Npi = 1000×11% = 110

expected frequency for  Montana

Fe = Npi = 1000×8% = 80

The Chi=square test statistics is;

x² = ∑ [ (Fo - Fe)² / Fe )

now we substitute

Given our Fo ( 450 in Washington, 340 in Oregon, 150 in Idaho, and 60 in Montana) from the question

x² =  (450-510)²/510 + (340-300)²/300 + (150-110)²/110 + (60-80)²/80

x²= 31.9376

so the chi-square test statistics is 31.9376

Now the chi-square critical value

first we compute for degree of freedom

d.f = k -1 = 4 - 1 = 3

So from the critical value table, degree of freedom 3 and significance level 0.05,

the chi-square critical value is 7.8147

Therefore chi-square test statistics 31.9376 is greater than the chi-square critical value 7.8147, so NULL HYPOTHESIS IS REJECTED at 5% significance.

There is sufficient evidence to warrant rejection  of the claim that the distribution of the sample agrees with the distribution of the state populations.

hope this !

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b segment jm is correct i think