Answer to Question #181089 in Statistics and Probability for Jii

Question

A research conducted on 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 decided 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

Let $p$ be the proportion of the population of coffee drinkers

$p_0=\frac{25}{100}=0.25,$ the proportion of employees that would rather drink coffee than soft drinks during break time.

$\hat{p}=\frac{28}{100}=0.28$, proportion that said they would rather drink coffee than soft drinks out of a sample of 125 ($n$ ).

To establish whether there is sufficient evidence to suggest that the coffee drinkers or the proportion of coffee drinkers ($p$) have increased since the company has decided to give free coffee during break time, we proceed as follows.

Step 1: We specify the null and alternative hypotheses: We use the notation $p$ which has already been introduced.

$H_0: p=0.25$ $H_a: p>0.25$

Step 2: Compute the test statistic.

$z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}=\frac{0.28-0.25}{\sqrt{\frac{(0.25)(0.75)}{125}}}=\frac{0.03}{0.0387}=0.7752.$

Step 3: Find the p-value.

From the standard normal table, the area below 0.7752=0.7823, so the p-value =1-0.7823=0.2177.

Step 4: Make a conclusion about the hypotheses at $\alpha=0.05$.

Since p-value=0.2177>0.05, the null hypothesis is not rejected implying that the result is not statistically significant.

Step 5: Report the conclusion in the context of the question.

Since the null hypothesis is not rejected, we can conclude that there is no sufficient evidence to suggest that the coffee drinkers have increased since the company decided to give free coffee during break time.