Question

Dropouts. Some people are concerned that new tougher standards and high-stakes tests adopted in many states have driven up the high school dropout rate. The National Center for Education Statistics reported that the high school dropout rate for the year 2004 was 10.3%. One school district whose dropout rate has always been very close to the national average reports that 210 of their 1782 high school students dropped out last year. Is this evi- dence that their dropout rate may be increasing

Answer 1

Answer:

There is evidence. It can be rejected, with 95% of confidence, that there is no dropout.

Step-by-step explanation:

The hypothesis for the school gives us a binomial random variable B(n = 1782, p = 10.3/100 = 0.103. This variable can be approximated to a Normal random variable with mean 1782*0.103 = 183.546 and a variance of 1782*0.103(1-0.103) = 164.64. Hence its standard deviation is √164.64 = 12.83

In order to irfer that the dropout is increasing or not, we can calculate the probability of obtaining a sample as the given to us for that school. Is the probability is very low, then we can confirm that the drop out increased.

Lets call X the dropout distribution of the school. X is approximately N(183.645, 12,83). If we standarize X, we obtain a variable Z with distribution Standard Normal N(0,1)

$Z = \frac{Z-183.546}{12.83}$

The values of the cummulative distribution of the standard normal, $\phi$ can be found in the attached file.

$P( X > 210) = P(\frac{X-183.546}{12.83} > \frac{210-183.546}{12.83} = 2.06) = P(W > 2.06)\\\\ = 1-\phi(2.06) = 1-0.9803 = 0.0197$

Since the probability of X being greater than 210 is quite low (less than 0.05), we can reject the hypothesis that there isnt a dropout with confidence 95%.