Question #263680


4 (a) In a random sample of 200 observations, we found the proportion of successes to be 48%. Estimate with 95% confidence the population proportion of successes.

(b) Repeat part (a) with n = 500.

(c) Repeat part (a) with n = 1000.

(d) Describe the effect on the confidence interval

estimate of increasing the sample size.


1
Expert's answer
2021-11-10T16:27:08-0500

(a) 95%CI=(0.481.960.48(10.48)200,0.48+1.960.48(10.48)200)=(0.4108,0.5492).95\%CI=(0.48-1.96\sqrt{\frac{0.48(1-0.48)}{200}},0.48+1.96\sqrt{\frac{0.48(1-0.48)}{200}})=(0.4108,0.5492).


(b) 95%CI=(0.481.960.48(10.48)500,0.48+1.960.48(10.48)500)=(0.4352,0.5238).95\%CI=(0.48-1.96\sqrt{\frac{0.48(1-0.48)}{500}},0.48+1.96\sqrt{\frac{0.48(1-0.48)}{500}})=(0.4352,0.5238).


(c) 95%CI=(0.481.960.48(10.48)1000,0.48+1.960.48(10.48)1000)=(0.4490,0.5110).95\%CI=(0.48-1.96\sqrt{\frac{0.48(1-0.48)}{1000}},0.48+1.96\sqrt{\frac{0.48(1-0.48)}{1000}})=(0.4490,0.5110).


(d) As the sample size increases, the confidence interval narrows.


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