Answer to Question #195991 in Statistics and Probability for sean

Question

B. Directions: Read and understand each problem carefully. In each of the following problems:

1. state the null and alternative hypothesis;

2. compute the test statistics value of population proportion;

3. determine the critical value and the rejection region; and

4. draw a conclusion.

1. A research conducted in a certain company last year showed that 25% of the employees would rather drink coffee than soft drinks during break time. The company has recently divided to give free coffee during break time. In the new research conducted this year, out of the 125 randomly sampled employees 28% said that they would rather drink coffee than soft drinks. At 0.05 level of significance, is there sufficient evidence to suggest that the coffee drinkers have increased since the company has decided to give free coffee during break time? 

Answer 1

Hypothesized Population Proportion $p_0=0.25$  

Favorable Cases $X=$

Sample Size $n=450$

Sample Proportion $\hat{p}=\dfrac{X}{n}=0.28$

Significance Level $\alpha=0.05$  

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

$H_0:p\leq0.25$

$H_1:p>0.25$

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 $\alpha=0.05,$ 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:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.28-0.25}{\sqrt{\dfrac{0.25(1-0.25)}{125}}}$$ $$=0.7746$$

Since it is observed that $z=0.7746<1.6449=z_c,$ 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 greater than $0.25,$ at the $\alpha=0.05$ significance level.

Using the P-value approach: The p-value is $p=0.219288,$ and since $p=0.219288>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is greater than $0.25,$ at the $\alpha=0.05$ significance level.

Therefore, there is not enough evidence to claim that the coffee drinkers have increased since the company has decided to give free coffee during break time, at the $\alpha=0.05$ significance level.