Questions: At "Stavros Restaurant" the owner is told by his chef that 25% of the customers order vegetarian food. The owner wants to check the validity of the chef's assertion so he checks a random sample of 20 orders, only to find 2 vegetarian orders. a) Is there evidence, at the 10% level of significance, that the proportion of vegetarian orders is lower than 25%? At "Mavros Restaurant" the owner is told by his waiters that 25% of the customers order vegetarian food. The owner wants to check the validity of the waiters' belief so he checks a random sample of 100 orders. b) Given that there are 31 vegetarian orders in the sample, use a distributional approximation, to test at the 5% level of significance, the belief of the waiters at "Mavros Restaurant". No credit will be given in part b) to answers which do not use a distributional approximation.

Solution Steps

The owner of Stavros Restaurant wants to test the chef's assertion that $25\%$ of the customers order vegetarian food. We set up the following hypotheses:

• Null Hypothesis ($H_0$): $p = 0.25$
• Alternative Hypothesis ($H_a$): $p < 0.25$

Using a sample of $n = 20$ orders, with $k = 2$ vegetarian orders, we calculate the sample proportion:

$$\hat{p} = \frac{k}{n} = \frac{2}{20} = 0.1$$

The test statistic is calculated as:

$$Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} = \frac{0.1 - 0.25}{\sqrt{\frac{0.25(1 - 0.25)}{20}}} = -1.5492$$

The corresponding p-value is $0.0607$.

At the $10\%$ significance level, the critical region is defined as $Z < -1.2816$. Since $-1.5492 < -1.2816$, we reject the null hypothesis.

The owner of Mavros Restaurant also wants to test the waiters' belief that $25\%$ of the customers order vegetarian food. The hypotheses are:

• Null Hypothesis ($H_0$): $p = 0.25$
• Alternative Hypothesis ($H_a$): $p \neq 0.25$

In this case, the sample size is $n = 100$ with $k = 31$ vegetarian orders. The sample proportion is:

$$\hat{p} = \frac{k}{n} = \frac{31}{100} = 0.31$$

The test statistic is calculated as:

$$Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} = \frac{0.31 - 0.25}{\sqrt{\frac{0.25(1 - 0.25)}{100}}} = 1.3856$$

The corresponding p-value is $0.1659$.

At the $5\%$ significance level, the critical region is defined as $Z < -1.96$ or $Z > 1.96$. Since $1.3856$ does not fall into the critical region, we fail to reject the null hypothesis.

Final Answer