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This question is incomplete, the complete question is;
Suppose an experiment is done with criminals released from prison in a certain state where the recidivism rate is 37%; that is, 37% of criminals return to prison within three years.
One hundred random prisoners are made to attend a "boot camp" for two weeks before their release, and it is hoped that "boot camp" will have a good effect. Suppose 32 of those prisoners return to prison within three years. The null hypothesis is that those attending boot camp have a recidivism rate of 37%.
a) what is p. sample proportion of successes?
b) what is p0, the hypothetical proportion of success under the null hypothesis
c) what is the value of test statistic
Answer:
a) sample proportion of successes is 0.32
b) the required value of p₀ is 0.37
c) the required test statistics value is -1.0356
Step-by-step explanation:
Given the data in the question
100 prisoners attend boot camp and 32 of them return to prison within three years
x = 32
n = 100
recidivism rate for the whole state = 37% = 0.37
a)
what is p. sample proportion of successes
p" = x / n
we substitute
p" = 32 / 100
p" = 0.32
Therefore, sample proportion of successes is 0.32
b)
what is p₀, the hypothetical proportion of success under the null hypothesis
given that;
the population recidivism rate for the whole state is 37%;
p₀ = 37%
p₀ = 0.37
Therefore, the required value of p₀ is 0.37
Null hypothesis H₀ : p₀ = 0.37
Alternative hypothesis Hₐ : p₀ ≠ 0.37
c)
what is the value of test statistic
Z = (p" - p₀) / √([tex]\frac{p_0(1-p_0)}{n}[/tex])
so we substitute
= (0.32 - 37) / √([tex]\frac{0.37(1-0.37)}{100}[/tex])
= -0.05 / √0.002331
= -0.05 / 0.04828
= -1.0356
Therefore, the required test statistics value is -1.0356