The restaurant manager is testing the bartender's ability to pour 45 mL of spirits correctly into a mixed drink. The manager has the bartender pour water into 12 shot glasses to test their ability to pour the correct amount of spirits:

48

45

44

43

46

47

42

46

47

45

47

49

Note: The data appears to be approximately normally distributed.

Test the bartender's ability to pour 45 mL at the 20% level of significance.

T-Distribution Table

**a.** Calculate the sample mean and standard deviation.

 x̄ = x̄ = 

Round to three decimal places if necessary

s=s= 

Round to three decimal places if necessary

**b.** Calculate the test statistic.

t=t= 

Round to three decimal places if necessary

**c.** Determine the critical value(s) for the hypothesis test.

 +

Round to three decimal places if necessary

**d.** Conclude whether to reject the null hypothesis or not based on the test statistic.

Reject

Fail to Reject




We are tasked with testing the bartender's ability to pour 45 mL of spirits using a hypothesis test at a 20% level of significance (α = 0.20). Since the data appears to be normally distributed and the sample size is small (n = 12), we will use the **t-distribution** for this test.

### Steps:

1. **State the hypotheses**:
   - Null hypothesis (\(H_0\)): The bartender pours exactly 45 mL on average, \( \mu = 45 \).
   - Alternative hypothesis (\(H_1\)): The bartender does not pour 45 mL on average, \( \mu \neq 45 \).

This is a **two-tailed** test because we are checking if the bartender pours either more or less than 45 mL.

### **Part a: Calculate the sample mean and standard deviation**

Given dataWe are tasked with testing the bartender's ability to pour 45 mL of spirits using a hypothesis test at a 20% level of significance (α = 0.20). Since the data appears to be normally distributed and the sample size is small (n = 12), we will use the **t-distribution** for this test.

### Steps:

1. **State the hypotheses**:
   - Null hypothesis (\(H_0\)): The bartender pours exactly 45 mL on average, \( \mu = 45 \).
   - Alternative hypothesis (\(H_1\)): The bartender does not pour 45 mL on average, \( \mu \neq 45 \).

This is a **two-tailed** test because we are checking if the bartender pours either more or less than 45 mL.

### **Part a: Calculate the sample mean and standard deviation**

Given 

48, 45, 44, 43, 46, 47, 42, 46, 47, 45, 47, 49

We will calculate:
- Sample mean \( \bar{x} \)
- Sample standard deviation \( s \)

Let's compute these now.### **Part a: Results**

- Sample mean \( \bar{x} = 45.750 \)
- Sample standard deviation \( s = 2.050 \)

### **Part b: Calculate the test statistic**

The test statistic for a t-test is calculated as:

\[
t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}
\]

Where:
- \( \bar{x} \) is the sample mean,
- \( \mu_0 = 45 \) mL (the target mean under the null hypothesis),
- \( s \) is the sample standard deviation,
- \( n = 12 \) is the sample size.

Now, let's compute the t-statistic.### **Part b: Result**

The test statistic \( t = 1.267 \).

### **Part c: Determine the critical value(s)**

Since this is a two-tailed test at a 20% level of significance (\( \alpha = 0.20 \)), we will look for the critical values that correspond to a total tail area of 0.20. This means each tail has an area of \( \alpha/2 = 0.10 \).

With \( n - 1 = 11 \) degrees of freedom, we can find the critical t-values from the t-distribution table or use an appropriate calculator. Let's calculate the critical values.### **Part c: Result**

The critical t-value for a two-tailed test with 11 degrees of freedom and \( \alpha = 0.20 \) is \( \pm 1.363 \).

### **Part d: Conclusion**

- The calculated test statistic \( t = 1.267 \) is **less than** the critical value of \( 1.363 \).
- Since \( |t| < 1.363 \), we **fail to reject** the null hypothesis.

### Conclusion:
At the 20% significance level, there is not enough evidence to conclude that the bartender is pouring incorrectly. Therefore, we fail to reject the null hypothesis, implying that the bartender is pouring an average of 45 mL of spirits correctly.

---

Do you want any further details or have any questions?  
Here are 5 related questions for practice:

1. What would happen if the significance level was changed to 5%?
2. How would a one-tailed test affect the critical value?
3. Can we conclude that the bartender pours exactly 45 mL based on this test?
4. How is the test statistic affected by a smaller sample size?
5. How would increasing the number of trials improve the power of this test?

**Tip**: Always verify whether your test is one-tailed or two-tailed when looking up critical values in t-distribution tables.

The restaurant manager is testing the bartender's ability to pour 45 mL of spirits correctly into a mixed drink. The manager has the bartender pour water into 12 shot glasses to test their ability to pour the correct amount of spirits:

48, 45, 44, 43, 46, 47, 42, 46, 47, 45, 47, 49

Note: The data appears to be approximately normally distributed.

Test the bartender's ability to pour 45 mL at the 20% level of significance.

Test the bartender's ability to pour 45 mL of spirits using hypothesis testing at a 20% significance level. This problem involves calculating the sample mean, standard deviation, and t-statistic for a two-tailed test using the t-distribution. Learn the statistical approach to hypothesis testing with small sample sizes.

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