Question 1181434: A random sample of 250 bottles of juice drink were taken and was
found to have an average content that is fess than the company’s claim
that each bottle contains 500 mL of juice drink. Suppose that an
appropriate test statistic revealed a value of -1.75, is there enough
evidence to support their claim at 95% confidence? Sketch the rejection
region and locate test statistic value.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! **1. State the hypotheses:**
* **Null hypothesis (H₀):** The average content of juice bottles is greater than or equal to 500 mL (µ ≥ 500).
* **Alternative hypothesis (H₁):** The average content of juice bottles is less than 500 mL (µ < 500).
**2. Determine the level of significance:**
* α = 0.05 (95% confidence level)
**3. Identify the test statistic:**
* z = -1.75 (given)
**4. Find the critical value:**
* Since this is a one-tailed test (less than), we look for the critical z-value that corresponds to α = 0.05 in the left tail of the standard normal distribution.
* Using a z-table or calculator, we find that the critical z-value is approximately -1.645.
**5. Sketch the rejection region:**
```
Rejection Region
/
/
/
/
/
---------------------------------------/------------------------------------>
-1.645 0
^
|
Test Statistic (z = -1.75)
```
**6. Make a decision:**
* Our calculated test statistic (z = -1.75) falls *within* the rejection region (z < -1.645).
* Therefore, we *reject* the null hypothesis.
**7. Conclusion:**
There is sufficient evidence at the 0.05 significance level to support the claim that the average content of juice bottles is less than 500 mL.
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