we believe that 42% of freshmen do not visit their counselors regularly. For this year, you would like to obtain a new sample to estimate th

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we believe that 42% of freshmen do not visit their counselors regularly. For this year, you would like to obtain a new sample to estimate the proportiton of freshmen who do not visit their counselors regularly. You would like to be 98% confident that your estimate is within 3.5% of the true population proportion. How large of a sample size is required

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Hồng Cúc 1 year 2021-08-21T22:12:28+00:00 1 Answers 0 views 0

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    2021-08-21T22:14:02+00:00

    Answer:

    A sample of 1077 is required.

    Step-by-step explanation:

    In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

    [tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

    In which

    z is the zscore that has a pvalue of [tex]1 – \frac{\alpha}{2}[/tex].

    The margin of error is of:

    [tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

    42% of freshmen do not visit their counselors regularly.

    This means that [tex]\pi = 0.42[/tex]

    98% confidence level

    So [tex]\alpha = 0.02[/tex], z is the value of Z that has a pvalue of [tex]1 – \frac{0.02}{2} = 0.99[/tex], so [tex]Z = 2.327[/tex].

    How large of a sample size is required?

    A sample size of n is required, and n is found when M = 0.035. So

    [tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

    [tex]0.035 = 2.327\sqrt{\frac{0.42*0.58}{n}}[/tex]

    [tex]0.035\sqrt{n} = 2.327\sqrt{0.42*0.58}[/tex]

    [tex]\sqrt{n} = \frac{2.327\sqrt{0.42*0.58}}{0.035}[/tex]

    [tex](\sqrt{n})^2 = (\frac{2.327\sqrt{0.42*0.58}}{0.035})^2[/tex]

    [tex]n = 1076.8[/tex]

    Rounding up:

    A sample of 1077 is required.

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