A marketing research company needs to estimate which of two medical plans its employees prefer. A random sample of n employees produced the

Question

A marketing research company needs to estimate which of two medical plans its employees prefer. A random sample of n employees produced the 95​% confidence interval (0.366 comma 0.506 )for the proportion of employees who prefer plan A. Identify the point estimate for estimating the true proportion of employees who prefer that plan.

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Anna 2 weeks 2021-10-08T10:50:52+00:00 2 Answers 0

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    0
    2021-10-08T10:52:08+00:00

    Answer:

    The point estimate for estimating the true proportion of employees who prefer that plan is 0.436

    Step-by-step explanation:

    A confidence interval has two bounds, a lower bound and an upper bound.

    A confidence interval is symmetric, which means that the point estimate used is the mid point between these two bounds, that is, the mean of the two bounds.

    In this problem, we have that:

    Lower bound: 0.366

    Upper bound: 0.506

    Point estimate:

    (0.366 + 0.506)/2 = 0.436

    The point estimate for estimating the true proportion of employees who prefer that plan is 0.436

    0
    2021-10-08T10:52:31+00:00

    Answer:

    \hat p = \frac{0.366+0.506}{2}= 0.436

    And the point of estimate for the proportion of  employees who prefer plan is on this case 0.436 or 43.6%.

    Step-by-step explanation:

    We define the parameter of interest as the proportion of employees who prefer plan p and we want to estimate this true parameter

    The confidence interval for the this proportion is given by the following formula:  

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

    After calculate the 95% confidence interval we got (0.366, 0.506)

    We can obtain the best estimate for the proportion of employees who prefer plan like this:

    \hat p =\frac{Upper +Lower}{2}

    Where Upper and Lower represent the limits for the confidence interval and replacing we got:

    \hat p = \frac{0.366+0.506}{2}= 0.436

    And the point of estimate for the proportion of  employees who prefer plan is on this case 0.436 or 43.6%.

    The margin of error can be estimated using the fact that this confidence interval is symmetrical and we got:

    ME=\frac{Upper-Lower}{2}= \frac{0.506-0.366}{2}=0.07

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