WineOrigin

In one study, 39 diners were given a free glass of Cabernet Sauvignon wine to accompany a French meal. Although the wine was identical, half the bottle labels claimed the wine was from California and the other half claimed it was from North Dakota. The following table summarizes the grams of entree and wine consumed during the meal. Did the patrons who thought that the wine was from California consume more?

| Wine Label | n | x̄ (grams) | s (grams) |
|-------------|----|------------|-----------|
| Entree | | | |
| California | 24 | 499.8 | 87.2 |
| North Dakota| 15 | 439.0 | 89.2 |
| Wine | | | |
| California | 24 | 100.8 | 23.3 |
| North Dakota| 15 | 110.4 | 9.0 |

1. What significance test should be performed?

A. Two-sample t-test
B. One-sample t-test
C. Paired t-test

2. What are the correct hypotheses to test this claim?

A. $ H_0: \mu_{\text{Cali}} = \mu_{\text{ND}} $ $ H_a: \mu_{\text{Cali}} < \mu_{\text{ND}} $
B. $ H_0: \mu_{\text{Cali}} = \mu_{\text{ND}} $ $ H_a: \mu_{\text{Cali}} > \mu_{\text{ND}} $
C. $ H_0: \mu_d = 0 $ $ H_a: \mu_d < 0 $
D. $ H_0: \mu_d = 0 $ $ H_a: \mu_d > 0 $

3. Find the test statistics and conservative degrees of freedom for both the amount of entree consumed and the amount of wine consumed. Use 2 decimals for the test statistics.

Entree:
$ t$-statistic = ??
degrees of freedom = ??

Wine:
$ t$-statistic = ??
degrees of freedom = ??

4. Which are true regarding the conclusion for the entree consumption data? Select all that apply: (3 are correct)

- $ p$-value < $\alpha$
- $ p$-value > $\alpha$
- statistically significant
- not statistically significant
- Reject $ H_0 $
- Fail to reject $ H_0 $

5. Which are true regarding the conclusion for the wine consumption data? Select all that apply: (3 are correct)

- $ p$-value < $\alpha$
- $ p$-value > $\alpha$
- statistically significant
- not statistically significant
- Reject $ H_0 $
- Fail to reject $ H_0 $

6. A manufacturer wishes to compare the mean lengths of items produced by two metal stamping processes. How many parts should be sampled to estimate the difference in means to within ±0.09 mm with 95% confidence? Previous studies indicate the standard deviation for both processes is about 1.25 mm.

- H0: uCali = uND (The mean consumption of patrons who thought the wine was from California is equal to the mean consumption of patrons who thought the wine was from North Dakota)

- Ha: uCali ≠ uND (The mean consumption of patrons who thought the wine was from California is not equal to the mean consumption of patrons who thought the wine was from North Dakota)

3) To find the test statistics and conservative degrees of freedom:

- For the amount of entree consumed:

- t-statistic = (x¯Cali - x¯ND) / sqrt((s^2Cali/nCali) + (s^2ND/nND))

- degrees of freedom = min(nCali - 1, nND - 1)

- For the amount of wine consumed:

- t-statistic = (x¯Cali - x¯ND) / sqrt((s^2Cali/nCali) + (s^2ND/nND))

- degrees of freedom = min(nCali - 1, nND - 1)

4) The true statements regarding the conclusion for the entree consumption data are:

- p-value > α (alpha)

- not statistically significant

- Fail to reject H0

5) The true statements regarding the conclusion for the wine consumption data are:

- p-value < α (alpha)

- statistically significant

- Reject H0

6) To estimate the difference in means to within ±0.09mm with 95% confidence, the sample size needed can be calculated using the formula:

- n = (z^2 * σ^2) / E^2

- where z is the z-value for the desired confidence level (for 95% confidence, z ≈ 1.96), σ is the standard deviation, and E is the desired margin of error.

- In this case, n = (1.96^2 * 1.25^2) / 0.09^2 ≈ 49.84

- Therefore, a sample size of at least 50 parts should be sampled to estimate the difference in means within ±0.09mm with 95% confidence.