Question

Suppose that several insurance companies conduct a survey. They randomly surveyed 350 drivers and found that 280 claimed to always buckle up. We are interested in the population proportion of drivers who claim to always buckle up. (.20) n = (.20) p' = (.20) The standard deviation for sp = (.20) The z value for a 95% confidence interval is = (.20) Construct a 95% confidence interval for the population proportion that claim to always buckle Fill in the blanks to clarify the following diagram. LL (lower limit) = UL (upper limit) =

Answer 1

Answer:

Lower limit = 0.76

Upper limit = 0.84

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 350

Number of drivers that buckle = 280

Formula:

$p' = \dfrac{x}{n} = \dfrac{280}{350} = 0.8$

$q' = 1-p' = 1 - 0.8 = 0.2$

The standard deviation for sp =

$=\sqrt{\displaystyle\frac{p'q'}{n}}\\\\=\sqrt{\dfrac{0.8\times 0.2}{350}} = 0.02138$

95% Confidence Interval:

$p' \pm z_{critical}(s_p)$

$z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.96$

Putting the values, we get,

$0.8 \pm 1.96(0.02138)\\= 0.8 \pm 0.0419048\\=(0.7580952, 0.8419048)\\\approx (0.76,0.84)$

Lower limit = 0.76

Upper limit = 0.84