Question

Water samples are taken from water used for cooling as it is being discharged from a power plant into a river. It has been determined that as long as the mean temperature of the discharged water is at most \(150^{\circ} \mathrm{F}\), there will be no negative effects on the river ecosystem. To investigate whether the plant is in compliance with regulations that prohibit a mean discharge water temperature above \(150^{\circ} \mathrm{F}\), a scientist will take 50 water samples at randomly selected times and will record the water temperature of each sample. She will then use a \(z\) statistic $$ z=\frac{\bar{x}-150}{\frac{\sigma}{\sqrt{n}}} $$ to decide between the hypotheses \(H_{0}: \mu=150\) and \(H_{a^{2}}\) \(\mu>150\), where \(\mu\) is the mean temperature of discharged water. Assume that \(\sigma\) is known to be 10 . a. Explain why use of the \(z\) statistic is appropriate in this setting. b. Describe Type I and Type II errors in this context. c. The rejection of \(H_{0}\) when \(z \geq 1.8\) corresponds to what value of \(\alpha ?\) (That is, what is the area under the \(z\) curve to the right of \(1.8 ?\) ) d. Suppose that the true value for \(\mu\) is 153 and that \(H_{0}\) is to be rejected if \(z \geq 1.8 .\) Draw a sketch (similar to that of Figure \(10.5\) ) of the sampling distribution of \(\bar{x}\), and shade the region that would represent \(\beta\), the probability of making a Type II error. e. For the hypotheses and test procedure described, compute the value of \(\beta\) when \(\mu=153\). f. For the hypotheses and test procedure described, what is the value of \(\beta\) if \(\mu=160 ?\) g. If \(H_{0}\) is rejected when \(z \geq 1.8\) and \(\bar{x}=152.8\), what is the appropriate conclusion? What type of error might have been made in reaching this conclusion?

Short answer

Use of z statistic is appropriate due to the Central Limit Theorem; Type I error is falsely claiming that the plant violates regulations, and Type II error is failing to detect an actual violation. The significance level is 0.0359. The values of \(\beta\) are 0.0170 for \(\mu = 153\) and nearly 0 for \(\mu = 160\). The test suggests that the power plant violates regulations, but there could be a potential Type I error.

Step-by-step solution

- Explaining use of z statistic

The z statistic is suitable in this case due to the Central Limit Theorem (CLT), which states that: if we draw a large enough sample, the distribution of the sample mean will be approximately normally distributed, independent of the population distribution, providing that the population's standard deviation is known and the sample size is large enough (N > 30). Here, the sample size is 50, which is > 30.

- Describing Type I and Type II errors

Type I Error: The null hypothesis \(H_0\) is true (i.e., the mean temperature of discharged water is 150) but is erroneously rejected. This means an incorrect claim would be made that the plant is violating the regulations. \n Type II Error: The alternate hypothesis \(H_a\) is true (i.e., the mean temperature of discharged water is >150) and yet the null hypothesis is not rejected. The error here is failing to detect a violation of regulations.

- Calculating value of alpha

The rejection area to the right of z = 1.8 on a z-distribution table corresponds to the significance level \(\alpha\). Looking up a Z-table for the value of 1.8, we get 0.9641. Therefore, \(\alpha = 1 - 0.9641 = 0.0359\).

- Sketching sampling distribution

The sampling distribution of \(\bar{x}\) would be a bell shaped curve centered at 150 (assuming \(H_0\) is true), with a standard deviation equal to \(\frac{\sigma}{\sqrt{n}} = \frac{10}{\sqrt{50}}\). For \(H_{a^{2}}\) (with \(\mu\) = 153), the curve is shifted to the right. The region to the right of z = 1.8 under the curve for \(H_{a^{2}}\) represents \(\beta\), the probability of making a Type II error.

- Computing value of beta when μ = 153

We first compute the corresponding z-score: \(z = \frac{153 - 150}{\frac{10}{\sqrt{50}}} = 2.12 \). Since the critical z = 1.8, we are interested in the area to the right of z = 2.12 under the standard normal curve. Looking up a Z-table, this corresponds to 1 - 0.9830 = 0.0170.

- Computing value of beta when μ = 160

The z-score is \(z = \frac{160 - 150}{\frac{10}{\sqrt{50}}} = 7.07 \). As 7.07 is quite large, it's safe to assume that the value of \(\beta\) will be nearly zero.

- Drawing conclusion

Since z = 1.8 is the decision boundary, and the computed z for \(\bar{x} = 152.8\) is \(z = \frac{152.8 - 150}{\frac{10}{\sqrt{50}}} = 1.988\) which is greater than 1.8, we reject \(H_0\). This suggests that the power plant is not complying with the regulations. However, we may have committed a Type I error, implying that although we concluded that the plant does not comply, in reality, it does.