Answer to Question #203138 in Statistics and Probability for yoonji

Question #203138

It is found out that 13% of students enrolled in College Algebra at a large university fail the first time they take it. A psychology professor claims that this could be reduced with a program of counseling, and so on. A random sample of 850 students enrolled in College Algebra take this program and 99 fail anyway. Is this a significant improvement at the 0.05 level?


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Expert's answer
2021-06-08T04:27:13-0400

Hypothesized Population Proportion p0=0.13p_0=0.13

Favorable Cases X=99X=99

Sample Size n=850n=850

Sample Proportion p^=998500.11647059\hat{p}=\dfrac{99}{850}\approx0.11647059

Significance Level α=0.05\alpha=0.05

The following null and alternative hypotheses for the population proportion needs to be tested:

H0:p0.13H_0:p\geq0.13

H1:p<0.13H_1:p<0.13

This corresponds to a left-tailed test, for which a z-test for one population proportion will be used.

Based on the information provided, the significance level is α=0.05,\alpha=0.05, and the critical value for a left-tailed test is zc=1.6449.z_c=-1.6449.

The rejection region for this left-tailed test is R{z:z<1.6449}R\{z:z<-1.6449\}

The z-statistic is computed as follows:


z=p^p0p0(1p0)nz=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}

=998500.130.13(10.13)8501.1729=\dfrac{\dfrac{99}{850}-0.13}{\sqrt{\dfrac{0.13(1-0.13)}{850}}}\approx-1.1729

Since it is observed that z=1.1729>1.6445=zc,z=-1.1729>-1.6445=z_c, it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion pp is less than 0.13,0.13, at the α=0.05\alpha=0.05 significance level.

Using the P-value approach:

The p-value is p=P(Z<1.1729)=0.120418,p=P(Z<-1.1729)=0.120418, and since p=0.0.120418>0.05=α,p=0.0.120418>0.05=\alpha, it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion pp is less than 0.13,0.13, at the α=0.05\alpha=0.05 significance level.


Therefore, there is not enough evidence to claim that this is a significant improvement at the 0.05 level.


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