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2 years ago

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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.

1 answer:

daser333 [38]2 years ago

4 0

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%

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Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

For this case we have the following info related to the time to prepare a return

$\mu =90 , \sigma =14$

And we select a sample size =49>30 and we are interested in determine the standard deviation for the sample mean. From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard deviation would be:

$\sigma_{\bar X} =\frac{14}{\sqrt{49}}= 2$

And the best answer would be

b. 2 minutes

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