Suppose that insurance companies did a survey. They randomly surveyed 400 drivers and found that 320 claimed they always buckle up. We are i

Question

Suppose that insurance companies did 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.

Compute the margin of error for a 95% confidence interval and construct this confidence interval for the population proportion who claim they always buckle up.
(i) State the margin of error. (Round your answers to four decimal places.)
(ii) State the confidence interval. (Round your answers to four decimal places.)

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Parker 1 week 2021-09-15T21:20:54+00:00 1 Answer 0

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    2021-09-15T21:22:52+00:00

    Answer:

    i) ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

    And replacing we got:

     ME= 1.96*\sqrt{\frac{0.8 (1-0.8)}{400}} =0.0392

    ii)  Lower = 0.8-0.0392= 0.7608

     Upper = 0.8 +0.0392= 0.8392

    Step-by-step explanation:

    Notation and definitions

    X=320 number of people that claimed always buckle up.

    n=400 random sample taken

    \hat p=\frac{320}{400}=0.8 estimated proportion of people that claimed always buckle up

    p true population proportion of people that claimed always buckle up

    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{p(1-p)}{n}})

    Solution to the problem

    Part i

    In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

    z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

    The confidence interval for the mean is given by the following formula:  

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

    The margin of error is given by:

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

    And replacing we got:

     ME= 1.96*\sqrt{\frac{0.8 (1-0.8)}{400}} =0.0392

    Part ii

    And the confidence interval would be given by:

     Lower = 0.8-0.0392= 0.7608

     Upper = 0.8 +0.0392= 0.8392

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