Question

The margarita is one of the most common tequila-based cocktails, made with tequila mixed with triple sec and lime or lemon juice, often served with salt on the glass rim. A common ratio for a margarita is 2:1:1, which includes 50% tequila, 25% triple sec, and 25% fresh lime or lemon juice. A manager at a local bar is concerned that the bartender uses incorrect proportions in more than 50% of margaritas. He secretly observes the bartender and finds that he used the correct proportions in only 10 out of 30 margaritas. Test if the manager’s suspicion is justified at α = 0.05. Let p represent incorrect population proportion.

Answer 1

Answer:

The null hypothesis was rejected.

Conclusion: The bartender uses incorrect proportions in more than 50% of margaritas.

Step-by-step explanation:

The hypothesis for this test can be defined as:

H₀: The bartender uses incorrect proportions in less than 50% of margaritas, i.e. p < 0.50.

Hₐ: The bartender uses incorrect proportions in more than 50% of margaritas, i.e. p > 0.50.

Given:

$\hat p=\frac{10}{30} =0.33\\n=30$

The test statistic is:

$z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}} }=\frac{0.33-0.50}{\sqrt{\frac{0.50(1-0.50)}{30}}} =-1.862$

Decision Rule:

If the p-value of the test statistic is less than the significance level, α = 0.05, then the null hypothesis is rejected.

The p-value of the test is:

$p-value=P(Z<-1.86) = 0.0314$

*Use the z-table.

The p-value = 0.0314 < α = 0.05.

The null hypothesis will be rejected.

Conclusion:

As the null hypothesis was rejected it can be concluded that the bartender uses incorrect proportions in more than 50% of margaritas.

Answer 2

Using the z-distribution, it is found that since the test statistic is less than the critical value for the left-tailed test, there is enough evidence to conclude that the manager's suspicion is correct.

At the null hypothesis, we test if the manager's suspicion is incorrect, that is, the proportion is correct in at least 50% of the margaritas.

$H_0: p \geq 0.5$

At the alternative hypothesis, we test if the suspicion is correct, that is, the proportion is less than 50%.

$H_1: p < 0.5$

The test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

In which:

• $\overline{p}$ is the sample proportion.
• p is the proportion tested at the null hypothesis.
• n is the sample size.

For this problem, the parameters are: $p = 0.5, n = 30, \overline{p} = \frac{10}{30} = 0.3333$

Then, the value of the test statistic is:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

$z = \frac{0.3333 - 0.5}{\sqrt{\frac{0.5(0.5)}{30}}}$

$z = -1.83$

The critical value for a left-tailed test, as we are testing if the mean is less than a value, with a significance level of 0.05, is $z^{\ast} = -1.645$.

Since the test statistic is less than the critical value for the left-tailed test, there is enough evidence to conclude that the manager's suspicion is correct.

A similar problem is given at https://brainly.com/question/23265899