Answer to Question #305497 in Statistics and Probability for marky boy

Question

The school registrar estimates that the dropout rate of freshmen high schools in Mindanao

is 20%. Last year, 45 freshmen from a random sample of 250 Mindanao freshmen high

school withdrew. At 𝛼 = 0.01, is there enough evidence to reject the registrar’s claim?

Answer 1

Hypotheses

H₀: p=0.2

H_a: p≠0.2

α = 0.01

Test statistic

 To use the Z score statistic for testing proportions, the condition that min(n*p, n*p(1-p))≥5 must be satisfied.

Given n=250 and p=0.2

min(250*0.2, 250*0.2*0.8) = min(50, 40) = 40

Since the condition is met, the Z test statistic is the appropriate test statistic for testing the hypothesis and is given by:

z = $\frac{p̂ - p}{\sqrt{p(1-p)/n}}$

where:

p̂ = sample proportion, p = hypothesized population proportion

n = sample size

Decision rule

From the Z normal distribution table, the critical value for a 99% significance level for two-tailed test is ±2.58. Therefore, we reject the null hypothesis if the computed test statistic Z ≤ -2.58 (lower critical region) or if Z ≥ 2.58 (upper critical region)

Computed test statistic

Let x be the number of freshmen in high school who drop out

Given that 45 out of 250 dropped out, then

Then x = 45 and sample size n=250

Then,

p̂ = x/n = 45/250 = 0.18

and p=0.8

Therefore

z =$\frac{p̂ - p}{\sqrt{p(1-p)/n}}$

= $\frac{0.18 - 0.2}{\sqrt{0.2(0.8)/250}}$

= -0.79

Decision

Since the computed test statistic Z= -0.79 > -2.58, we fail to reject the null hypothesis.

Conclusion

At 1% level of significance, there is insufficient evidence to reject the registrar’s claim that the school dropout rate of freshmen in high schools in Mindanao is 20%.

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