Answer to Question #203138 in Statistics and Probability for yoonji

Question

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?

Answer 1

Hypothesized Population Proportion $p_0=0.13$

Favorable Cases $X=99$

Sample Size $n=850$

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

Significance Level $\alpha=0.05$

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

$H_0:p\geq0.13$

$H_1:p<0.13$

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 $\alpha=0.05,$ and the critical value for a left-tailed test is $z_c=-1.6449.$

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

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}$$

$$=\dfrac{\dfrac{99}{850}-0.13}{\sqrt{\dfrac{0.13(1-0.13)}{850}}}\approx-1.1729$$

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

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

Using the P-value approach:

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

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

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