Question

Suppose that the insurance companies did do a survey. They randomly surveyed $400$drivers and found that $320$claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up.

a. i. x = __________ ii. n = __________ iii. p′ = __________

b. Define the random variables $X$and $P\text{'}$, in words.

c. Which distribution should you use for this problem? Explain your choice.

d. Construct a $95\%$confidence interval for the population proportion who claim they always buckle up. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound.

e. If this survey were done by telephone, list three difficulties the companies might have in obtaining random results.

Short answer

a). The quantity of successes is $x=320$.

The sample size is $n=400$drivers.

The purpose estimate for truth population proportion is $\widehat{p}=0.8$.

b. stochastic variable $X$is the amount of$"\left\{successes\right\}"$where drivers who claim they always buckle up.

$\widehat{p}$is the percentage of drivers sampled who claim they always buckle up.

c. We use the conventional distribution $\mathcal{N}\left(0.8,\sqrt{\frac{0.8\times 0.2}{400}}\right)$.

d. i) A $95\%$confidence interval for the population proportion is $0.761\le p\le 0.839$and $EBM=0.039$

ii)

Step-by-step solution

Successes  (a),(b) and (c)

a) - The whole number of successful attempts is $x=320$.

- There are $n=400$ drivers within the sample.

- The population mean percent information is $\widehat{p}=\frac{x}{n}=\frac{320}{400}=0.8$

b) Component at unexpected times Average amount of "successes" inside which users claim to carry on tight is $x$.

The proportion of drivers polled who indicate their constantly buckle up is $\widehat{p}$.

c) c) Given $\widehat{p}=0.8$and $n=400$, the dispersion we must follow that compute a little is

$\mathcal{N}\left(0.8,\sqrt{\frac{0.8\times 0.2}{400}}\right)$

Confidence Interval (d)

d) If the percent of sightings during a randomized subset $n$ that relate to a category of things is $\widehat{p}$, an estimate $100(1-\alpha )\mathrm{\%}$ significance level here on proportion $p$ of such populace that belong to the current group is .

$\widehat{p}-z_{\frac{\alpha }{2}}\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\le p\le \widehat{p}+z_{\frac{\alpha }{2}}\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

localid="1649850739294" $\frac{\alpha }{2}=\frac{1-0.95}{2}$

localid="1649850746722" $z_{\frac{\alpha }{2}}=1.96$

localid="1649850753572" $\begin{array}{r}0.8-1.96\sqrt{\frac{0.8(1-0.8)}{400}}\le p\le 0.8+1.96\sqrt{\frac{0.8(1-0.8)}{400}}\\ \end{array}$

localid="1649850761120" $0.8-0.039\le p\le 0.8+0.039$

localid="1649850768823" $0.761\le p\le 0.839$

This problem's graph is as follows:

Error bound

- The upper limit of the error limitation is

$EBM=0.039$

Is that if poll were conducted over the phone, the businesses might face three challengesin an exceedingly achieving random results:

- obtain results from people in a same town,

- obtain results from people of comparable ages,

- obtain inaccurate findings.