Question 1177611: In a study of factors affecting hypnotism, visual analogue scale (VAS) sensory ratings
were obtained for 16 subjects. For these sample ratings, the mean is 8.33 with a standard
deviation of 1.96. At
0.05
level of significance test the claim that this sample comes
from a population with a mean rating of less than 9.00.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! **1. State the Hypotheses**
* Null Hypothesis (H0): The population mean rating is equal to 9.00 (µ = 9.00)
* Alternative Hypothesis (H1): The population mean rating is less than 9.00 (µ < 9.00)
**2. Determine the Test Statistic**
Since the population standard deviation is unknown, we'll use a t-test. The test statistic is calculated as:
```
t = (x̄ - µ) / (s / √n)
```
Where:
* x̄ = sample mean (8.33)
* µ = hypothesized population mean (9.00)
* s = sample standard deviation (1.96)
* n = sample size (16)
Plugging in the values:
```
t = (8.33 - 9.00) / (1.96 / √16)
t ≈ -1.36
```
**3. Find the P-value**
We need to find the P-value associated with this t-statistic. Since the alternative hypothesis is µ < 9.00 (a one-tailed test), we want the area to the left of t = -1.36 on a t-distribution with 15 degrees of freedom (df = n - 1 = 15).
Using a t-table or calculator, the P-value is approximately 0.096.
**4. Compare P-value to Significance Level (α)**
* α = 0.05 (given)
* P-value (0.096) > α (0.05)
**5. Decision**
Since the P-value is greater than α, we fail to reject the null hypothesis.
**6. Conclusion**
There is not enough evidence at the 0.05 level of significance to support the claim that the sample comes from a population with a mean rating of less than 9.00.
**Interpretation**
While the sample mean (8.33) is less than 9.00, the difference is not statistically significant. The observed difference could be due to random chance.
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