Answer to Question #306160 in Statistics and Probability for nixx

Question

A certified public aacountant claims that more than 30% of all accountants advertise. A sample of 400 accounts in Rizal province showed that 112 had used some form of advertising. At a a=0.01,is there enough evidence to support the claim? (Use p-value method).

Solution:

Step 1:State the hypothesis.

H0:

H1:

Step 2:Compute for the value of one sample t test.zcomputed=

Step 3: Compute for th3 p-value.

Step 4:Decision rule:

Step 5: Conclusion

Answer 1

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

$H_0:p\le 0.3$

$H_1:p>0.3$

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

Based on the information provided, the significance level is $\alpha = 0.01,$ and the critical value for a right-tailed test is $z_c = 2.3263.$

The rejection region for this right-tailed test is $R = \{z: z > 2.3263\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{112/400-0.3}{\sqrt{\dfrac{0.3(1-0.3)}{400}}}=-0.87287$$

Using the P-value approach:

The p-value is $p=P(Z>-0.87287)=0.808633,$ and since $p=0.808633>0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is greater than $0.3,$ at the $\alpha = 0.01$significance level.