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α = 0.01
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:
p̂ = sample proportion, p = hypothesized population proportion
n = sample size
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
p̂ = x/n = 45/250 = 0.18
Since the computed test statistic Z= -0.79 > -2.58, we fail to reject the null hypothesis.
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%.