An education firm measures the high school dropout rate as the percentage of 16 to 24 year olds who are not enrolled in school and have not

Question

An education firm measures the high school dropout rate as the percentage of 16 to 24 year olds who are not enrolled in school and have not yet earned a high school credential. One year, the high school dropout rate was 6.1%. The next year a polling company took a survey of 1000 people between the ages of 16 and 24 and found that 56 of them are high school dropouts. The polling company would like to determine whether the dropout rate has decreased. The value of the test statistic is

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Emery 15 hours 2021-09-15T06:11:35+00:00 1 Answer 0

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    2021-09-15T06:12:59+00:00

    Answer:

    The value of the test statistic z = -0.6606

    Step-by-step explanation:

    From the question we are told that

          The high dropout rate is \mu  = 6.1%  = 0.061

          The  sample size is  n = 1000

           The  number of dropouts x = 56

           The probability of having a dropout in 1000 people  \= x  = \frac{56}{1000}  =  0.056

    Now setting up Test Hypothesis

     Null  H_o :  p = 0.061

     Alternative  H_a : p < 0.061        

    The Test statistics is mathematically represented as

                     z = \frac{\= x - p}{\sqrt{\frac{p(1 -p)}{n} } }

    substituting values

                   z = \frac{0.056 - 0.061}{\sqrt{\frac{0.061(1 -0.061)}{1000} } }

                   z = -0.6606

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