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) =

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ChiKesselman ChiKesselman · Answer 1 · 2020-02-05T09:52:52+01:00

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:

[tex]p' = \dfrac{x}{n} = \dfrac{280}{350} = 0.8[/tex]

[tex]q' = 1-p' = 1 - 0.8 = 0.2[/tex]

The standard deviation for sp =

[tex]=\sqrt{\displaystyle\frac{p'q'}{n}}\\\\=\sqrt{\dfrac{0.8\times 0.2}{350}} = 0.02138[/tex]

95% Confidence Interval:

[tex]p' \pm z_{critical}(s_p)[/tex]

[tex]z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.96[/tex]

Putting the values, we get,

[tex]0.8 \pm 1.96(0.02138)\\= 0.8 \pm 0.0419048\\=(0.7580952, 0.8419048)\\\approx (0.76,0.84)[/tex]

Lower limit = 0.76

Upper limit = 0.84