A paper reported a (1−α) confidence interval for the proportion of voters is (0.561,0.599) based on a sample of 2,056 people. However, the paper omitted the value of α. If you want to test the hypothesis that the proportion of voters is greater than 65% at 1% significance, find zcalc value for this problem? Please report your answer to 2 decimal places.

Question

ka1leE · Accepted Answer

Given data, Total number of people is 2056. The proportion of voters is (0.561, 0.599) Step 1 This is a symmetrical CI. Hence the sample proportion is expressed as, p^=0.561+0.5992 =0.58 Length of CI is expressed as difference between given proportion. Margin of error is expressed as, Margin of error(E) =12× length of CI =12×(0.599−0.561) =12×0.038 =0.019 Step 2 Hence the CI of the given proportion with margin of error is, p^+E=0.58+0.019=0.599 p^−E=0.58−0.019=0.561 Hence, the proportions are (0.599,0.561) The expression for standard deviation is, S.E=p^(1−p^)n =0.58×0.422056 =0.24362056 =0.00011848 =0.01088 Step 3 As know that, Margin of error (E)=zα2×S.E Hence, the value of z is, zα2=ES.E =0.0190.01088 ≈1.7463 Thus α2=P(Z&amp;lt;−1.7463) α2=0.0274 α=0.0548 Hence the confidence level is (1−α)=0.9452. Step 4 Now, to find null hypothesis: H0:p=65 H0:p&amp;gt;65 Here, sample proportion is 0.58,n=2056 and claimed proportion is 0.65 at significance level α=0.01. Hence the standard deviation at 0.65 claimed proportion, S.E=claimed proportion(1-claimed proportion)n =0.65×0.352056 =0.22752056 ≈0.0105 So the value of zcalc is: zcalc=p^−0.650.0105 =0.58−0.650.0105 =−0.070.0105 =−6.6667 Hence, the value of z is -6.67.

Answered question

Answer & Explanation

New Questions in College Statistics