4. (15 marks) A researcher is interested in estimating the proportion of populat

Question

4. (15 marks) A researcher is interested in estimating the proportion of population p that supports political party A in the election. Suppose that population size N -300,000 and he plans to draw a SRSWOR of size n from the population. (a) If the researcher wants his proportion estimator p to achieve the precision with tolerance level e 0.05 and risk a0.1. Estimate the minimal sample size needed for this estimation. (2005 1.65) Suppose the researcher chooses the sample size estimated from (a), and he finds that the sample proportion p 07 Construct a 90% confidence interval for population proportion (b)

Explanation / Answer

Here we have given population size (N) = 300,000

Level of significance (alpha) = 0.1

Precision with tolerance level (e) = 0.05

Zc = 1.65

We have to find sample size here.

Sample size we can calculate by using formula,

n = 1/4 * (Zc / e)^2

= 1/4 * (1.65/0.05)^2

= 272.25 which is approximately equal to 272

Now p^ = 0.7

n = 272

We have to find 90% confidence interval for p.

90% confidence interval for p is ,

p^ - E < p < p^ + E

where E is margin of error.

E = Zc * SE(p^)

Zc is critical value for normal distribution.

Zc = 1.65 (given)

SE(p^) = sqrt [(1-n/N) *(p^q^)/n-1]

q^ = 1 - p^

SE(p^) = sqrt [(1- 272 / 300,000) * (0.7*0.3) / 272 - 1 ]

= sqrt [0.000774]

= 0.028

E = 1.65*0.028 = 0.046

lower limit = p^ - E = 0.7 - 0.046 = 0.654

upper limit = p^ + E = 0.7 + 0.046 = 0.746

90% confidence interval for population proportion is (0.654, 0.746).

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