Answer:
The lower bound of a 99% C.I for the proportion of defectives = 0.422
Step-by-step explanation:
From the given information:
The point estimate = sample proportion [tex]\hat p[/tex]
[tex]\hat p = \dfrac{x}{n}[/tex]
[tex]\hat p = \dfrac{55}{100}[/tex]
[tex]\hat p[/tex] = 0.55
At Confidence interval of 99%, the level of significance = 1 - 0.99
= 0.01
[tex]Z_{\alpha/2} =Z_{0.01/2} \\ \\ = Z_{0.005} = 2.576[/tex]
Then the margin of error [tex]E = Z_{\alpha/2} \times \sqrt{\dfrac{\hat p(1-\hat p)}{n}}[/tex]
[tex]E = 2.576 \times \sqrt{\dfrac{0.55(1-0.55)}{100}}[/tex]
[tex]E = 2.576 \times \sqrt{\dfrac{0.2475}{100}}[/tex]
[tex]E = 2.576 \times0.04975[/tex]
E = 0.128156
E ≅ 0.128
At 99% C.I for the population proportion p is: [tex]\hat p - E[/tex]
= 0.55 - 0.128
= 0.422
Thus, the lower bound of a 99% C.I for the proportion of defectives = 0.422