Mathematics, 12.12.2019 05:31 ryleigh8211

Asample survey is designed to estimate the proportion of sports utility vehicles being driven in the state of california. a random sample of 500 registrations are selected from a department of motor vehicles database, and 48 are classified as sports utility vehicles. (a) use a 95% confidence interval to estimate the proportion of sports utility vehicles in california. (round your answers to three decimal places.) to (b) how can you estimate the proportion of sports utility vehicles in california with a higher degree of accuracy? (hint: there are two answers. select all that apply.) increase za/2 by increasing the confidence coefficient conduct a non-random sample increase the sample size n decrease the sample size n decrease za/2 by increasing the confidence coefficient decrease za/2 by decreasing the confidence coefficient increase za/2 by decreasing the confidence coefficient you may need to use the appropriate appendix table or technology to answer this question.

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Answer from: dsperez201938

a) The 95% confidence interval would be given (0.070;0.121).

b) "increase the sample size n"

"decrease za/2 by decreasing the confidence"

Step-by-step explanation:

Notation and definitions

X=48 number of vehicles classified as sports utility.

n=500 random sample taken

$\hat p=\frac{48}{500}=0.096$ estimated proportion of vehicles classified as sports utility vehicles.

true population proportion of vehicles classified as sports utility vehicles.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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 population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

(a) Use a 95% confidence interval to estimate the proportion of sports utility vehicles in California. (Round your answers to three decimal places.)

The confidence interval would be given by this formula :

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.5=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.096 - 1.96 \sqrt{\frac{0.096(1-0.096)}{500}}=0.070$

$0.096 + 1.96 \sqrt{\frac{0.096(1-0.096)}{500}}=0.121$

And the 95% confidence interval would be given (0.070;0.121).

We are confident (95%) that about 7.0% to 12.1% of vehicles in California are classified as sports utility .

(b) How can you estimate the proportion of sports utility vehicles in California with a higher degree of accuracy? (HINT: There are two answers. Select all that apply.)

For this case we just have two ways to increase the accuracy one is "increase the sample size n" since if we have a larger sample size the estimation would be more accurate. And the other possibility is "decrease za/2 by decreasing the confidence" because if we decrease the confidence level the interval would be narrower and accurate

Answer from: Quest

three-fourths of an hour

36÷48=0.75

0.75= $\frac{3}{4}$

Answer from: Quest

each side has a different length.

step-by-step explanation:

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Asample survey is designed to estimate the proportion of sports utility vehicles being driven in the...