In a recent poll, 370 people were asked if they liked dogs, and 7% said they did. Find the margin of error of this poll, at the 90% confiden

Question

In a recent poll, 370 people were asked if they liked dogs, and 7% said they did. Find the margin of error of this poll, at the 90% confidence level. Give your answer to three decimals.

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1 week 2021-10-11T18:51:54+00:00 1 Answer 0

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    2021-10-11T18:53:33+00:00

    Answer:

     ME=1.64\sqrt{\frac{0.07(1-0.07)}{370}}=0.0218

    Step-by-step explanation:

    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”.  

    p represent the real population proportion of interest

    \hat p=0.07 represent the estimated proportion for the sample

    n=370 is the sample size required (variable of interest)

    z represent the critical value for the margin of error

    The population proportion have the following distribution  

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

    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 90% of confidence, our significance level would be given by \alpha=1-0.90=0.10 and \alpha/2 =0.05. And the critical value would be given by:  

    z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64  

    The margin of error for the proportion interval is given by this formula:  

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

    And replacing into formula (a) the values provided we got:

     ME=1.64\sqrt{\frac{0.07(1-0.07)}{370}}=0.0218

    The margin of error on this case would be ME=0.0218

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