A random sample of 100 students shows that 12 are business students

Ans 1:

i) Let X = the number of Business students .

ii)To calculate the confidence interval, we must find p′, q′ , and EBP(Margin of Error) .

iii)n =100, x = the number of successes = 12 ,       p’ = x/n = 12/ 100 = 0.12

iv) p′ = 0.12 is the sample proportion; this is the point estimate of the population proportion.

v) q′ = 1 – p′ = 1 – 0.12 = 0.88

vi)Since CL = 0.99, then α = 1 – CL = 1 – 0.99 = 0.01   

vii) α /2 = 0.005 ,   Then Zα/2 = Z0.005 = 2.576

viii)As we recollect, that the area to the right of Z_0.005 is 0.005 and the area to the left of Z_0.005 is 0.995 (1 – 0.005) . The above can be done by using a Standard Normal probability table.

ix)EBP/ME = (Zα/2)(√p′q′ / n) = (2.576 ) √ (0.12)(0.88) / 100 = 0.0837

p’ − EBP = 0.12 − 0.0837 = 0.0363

p′ + EBP = 0.12 + 0.0837 = 0.2037

The 99% confidence interval for the true binomial population proportion is ( p′ – EBP, p′ + EBP) = (0.0363, 0.2037).

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