According to statistics reported on CNBC, a surprising number of motor vehicles are not covered by insurance (CNBC, February 23, 2006). Sample results, consistent with the CNBC report, showed 46 of 200 vehicles were not covered by insurance.
a. What is the point estimate of the proportion of vehicles not covered by insurance?
b. Develop a 95% confidence interval for the population proportion.

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umohduke14 umohduke14 · Answer 1 · 2019-10-23T20:10:18+02:00

Answer: a) 0.23. b) 0.1716 to 0.2884

Step-by-step explanation:

A.)

Point estimate is just the proportion of 46/200 = 23/100 = 0.23

B.)

95 % percent confidence interval is from 2.5 % to 97.5 % leaving ninety five percent in the middle

P(z<Z) = 0.975

Z = 1.96

mean = 0.23

standard deviation of a proportion = sqrt ( p*(1-p) ) = sqrt (0.23*(1- 0.23)) = 0.420832508

C1 (lower) = mean - Z_crit*standard deviation/ sqrt(N)

C1 (Lower) = 0.23 - 1.96*0.420832508/sqrt(200) = 0.17167559

C2 (High) = 0.23 +1.96*0.420832508/sqrt(200) = 0.28832441

Furthermore,

If the exact values are from .0.17167559 to 0.28832441 , then from 0.1717 to 0.2883 includes slightly less than 95 %

while 0.1716 to 0.2884 includes slightly more than 95%

topeadeniran2 topeadeniran2 · Answer 2 · 2022-03-25T14:53:33+01:00

The sample proportion is 0.23 and the confidence interval is (0.172, 0.288).

From the information given, the sample proportion will be:

p = x/n = 46/200 = 0.23

For 95% confidence interval, the critical value of z is 1.96. Therefore, the confidence interval for the population proportion will be:

= 0.23 + 1.96[✓(0.23(1 - 0.23)] / ✓200

= 0.288

Also, 0.23 - 1.96[✓(0.23(1 - 0.23)] / ✓200

= 0.058

Therefore, the confidence interval is (0.172, 0.288).

Learn more about confidence interval on:

https://brainly.com/question/25779324