SOLUTION: The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let μ denote the true average reflectometer reading for a new type of paint under c

Question 1186172: The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let μ denote the true average reflectometer reading for a new type of paint under consideration. A test of H_0: μ = 20 vs. H_A: μ > 20 will be based on a random sample of size n from a normal population distribution.
a. Suppose z=3.3. Calculate the p-value for the hypotheses listed above. Round your answer to four decimal places.
b. Suppose z=1.6. Calculate the p-value for the hypotheses listed above. Round your answer to four decimal places.
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Here's how to calculate the p-values for the given hypothesis test:
**Understanding the Problem**
We are conducting a one-tailed (right-tailed) z-test. Our null hypothesis is that the average reflectometer reading is equal to 20 (μ = 20), and our alternative hypothesis is that it is greater than 20 (μ > 20). The p-value represents the probability of observing a test statistic as extreme as, or more extreme than, the one calculated (in this case, a z-score), *assuming the null hypothesis is true*.
**a. z = 3.3**
1. **Find the area to the right of z:** Since this is a right-tailed test, the p-value is the area under the standard normal curve to the *right* of the calculated z-score.
2. **Use a z-table or calculator:** Look up the area corresponding to z = 3.3. Most z-tables give the area to the *left*. If that's the case, subtract the value you find from 1 to get the area to the right.
P(z > 3.3) ≈ 0.0005
**b. z = 1.6**
1. **Find the area to the right of z:** Again, since it is a right-tailed test, the p-value is the area to the right of the given z-score.
2. **Use a z-table or calculator:** Look up the area corresponding to z = 1.6.
P(z > 1.6) ≈ 0.0548
**Answers:**
* a. p-value ≈ 0.0005
* b. p-value ≈ 0.0548

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