An automatic coffee machine dispenses cups of coffee whose volume per cup varies normally with the mean mu = 10 oz. A quality-control researcher randomly selects 8 cups of coffee from the machine and finds that in this sample the mean volume is 9.92 oz. and the standard deviation is 0.23 oz. a. Do these data provide enough evidence to conclude that the mean volume per cup is below the target level? Which of the following two outputs represents the correct way to conduct this test? One-Sample T Test of mu = 10 vs < 10 95% Upper SE Me an 0.08132 N Mean 9.92000 StDev Bound T 8 0.23000 10.07406 -0.98 0.179 Output A: One-Sample Z Test of mu = 10 vs < 10 The assumed standard deviation = 0.23 95% Upper Bound 10.05376 Mean SE Me an Output B: 8 0.163 9.92000 0.08132 -0.98 O Output A O Output B b. Which of the following represents the correct conclusion we can make based on the output you selected in the previous problem (and at the usual significance level of .05)? O The data provide enough evidence to reject H, and to conclude that the mean volume per cup is lower than the target level of 10 oz. O The data provide enough evidence to accept H, and to conclude that the mean volume per cup is at the target level of 10 oz. O The data do not provide enough evidence to reject Ho, so we accept it, and conclude that the mean volume per cup is at the target level of 10 oz. O The data do not provide enough evidence to reject Ho, nor to conclude that the mean volume per cup is lower than the target level of 10 oz.
An automatic coffee machine dispenses cups of coffee whose volume per cup varies normally with the mean mu = 10 oz. A quality-control researcher randomly selects 8 cups of coffee from the machine and finds that in this sample the mean volume is 9.92 oz. and the standard deviation is 0.23 oz. a. Do these data provide enough evidence to conclude that the mean volume per cup is below the target level? Which of the following two outputs represents the correct way to conduct this test? One-Sample T Test of mu = 10 vs < 10 95% Upper SE Me an 0.08132 N Mean 9.92000 StDev Bound T 8 0.23000 10.07406 -0.98 0.179 Output A: One-Sample Z Test of mu = 10 vs < 10 The assumed standard deviation = 0.23 95% Upper Bound 10.05376 Mean SE Me an Output B: 8 0.163 9.92000 0.08132 -0.98 O Output A O Output B b. Which of the following represents the correct conclusion we can make based on the output you selected in the previous problem (and at the usual significance level of .05)? O The data provide enough evidence to reject H, and to conclude that the mean volume per cup is lower than the target level of 10 oz. O The data provide enough evidence to accept H, and to conclude that the mean volume per cup is at the target level of 10 oz. O The data do not provide enough evidence to reject Ho, so we accept it, and conclude that the mean volume per cup is at the target level of 10 oz. O The data do not provide enough evidence to reject Ho, nor to conclude that the mean volume per cup is lower than the target level of 10 oz.
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
Transcribed Image Text:### Assessing Mean Volume of Coffee Dispensed by an Automatic Machine
**Context:**
An automatic coffee machine dispenses cups of coffee with a normally distributed volume per cup around a mean (μ) of 10 ounces (oz.). A quality-control researcher randomly selects 8 cups of coffee from the machine and finds that in this sample, the mean volume is 9.92 oz. with a standard deviation of 0.23 oz.
**Question (a):**
Do these data provide enough evidence to conclude that the mean volume per cup is below the target level of 10 oz.?
**Outputs Provided for Statistical Testing:**
- **Output A: One-Sample T Test**
```
One-Sample T
Test of mu = 10 vs < 10
N Mean StDev SE Mean 95% Upper Bound T P
8 9.9200 0.2300 0.08132 10.07406 -0.98 0.179
```
- **Output B: One-Sample Z Test**
```
One-Sample Z
Test of mu = 10 vs < 10
The assumed standard deviation = 0.23
N Mean SE Mean 95% Upper Bound Z P
8 9.9200 0.08132 10.05376 -0.98 0.163
```
**Selecting the Correct Output method:**
The outputs provided show the results of two different tests—a T-test and a Z-test. The appropriate test in this scenario, given the small sample size (n = 8) and an unknown true population standard deviation, should be the one-sample T-test.
- Select **Output A**
---
**Question (b):**
Based on the selected output, which conclusion is correct using the usual significance level of 0.05?
- The hypothesis setup is:
- Null hypothesis (\( H_0 \)): μ = 10 oz.
- Alternative hypothesis (\( H_1 \)): μ < 10 oz.
Since the P-value in Output A from the T-test is 0.179, which is greater than the significance level of 0.05, the null hypothesis cannot be rejected.
- **Conclusion:**
- The data do not provide enough
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