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LiRa [457]
3 years ago
5

The Institute of Education Sciences measures the high school dropout rate as the percentage of 16- through 24-year-olds who are

not enrolled in school and have not earned a high school credential. In 2009, the high school dropout rate was 8.1%. A polling company recently took a survey of 1000 people between the ages of 16 and 24 and found 6.5% of them are high school dropouts. The polling company would like to determine whether the proportion of dropouts has changed from the historical value of 0.081. Use the 5% significance level. Set up the null and alternative hypotheses, calculate the Test Statistic, p-value, and write your conclusion in a sentence
Mathematics
1 answer:
STALIN [3.7K]3 years ago
7 0

Answer:

We conclude that the proportion of dropouts has changed from the historical value of 0.081.

Step-by-step explanation:

We are given that in 2009, the high school dropout rate was 8.1%. A polling company recently took a survey of 1000 people between the ages of 16 and 24 and found 6.5% of them are high school dropouts.

The polling company would like to determine whether the proportion of dropouts has changed from the historical value of 0.081.

<em>Let p = proportion of school dropouts rate</em>

SO, <u>Null Hypothesis,</u> H_0 : p = 0.081   {means that the proportion of dropouts has not changed from the historical value of 0.081}

<u>Alternate Hypothesis</u>, H_A : p \neq 0.081   {means that the proportion of dropouts has changed from the historical value of 0.081}

The test statistics that will be used here is <u>One-sample z proportion statistics</u>;

             T.S.  = \frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }  ~ N(0,1)

where, \hat p  = sample proportion of high school dropout rate = 6.5%

            n = sample of people = 1000

So, <u><em>test statistics</em></u>  =   \frac{0.065-0.081}{{\sqrt{\frac{0.065(1-0.065)}{1000} } } } } 

                               =  -2.05

<u>Also, P-value is given by the following formula;</u>

         P-value = P(Z < -2.05) = 1 - P(Z \leq 2.05)

                                              = 1 - 0.97982 = <u>0.0202</u> or 2.02%

<em>Now at 5% significance level, the z table gives critical values between -1.96 and 1.96 for two-tailed test. Since our test statistics does not lies within the range of critical values of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.</em>

Therefore, we conclude that the proportion of dropouts has changed from the historical value of 0.081.

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