It is desired to test $H _ { 0 } : \mu = 55$ against $H _ { \mathrm { a } } : \mu

Question

It is desired to test $H _ { 0 } : \mu = 55$ against $H _ { \mathrm { a } } : \mu < 55$ using $\alpha = .10$. The population in question is uniformly distributed with a standard deviation of 15 . A random sample of 49 will be drawn from this population. If $\mu$ is really equal to 50 , what is the power of this test?
A) $.3531$
B) $.8531$
C) $.2938$
D) .1469

Quizplus · Accepted Answer

The power of a test is the probability of correctly rejecting the null hypothesis when it is false. To calculate the power, we first determine the critical value using the significance level ($\alpha$) and then calculate the probability of observing a test statistic that leads to rejecting $H_0$ given that the true mean ($\mu$) is 50. Given $\alpha = .10$ for a left-tailed test, the Z critical value (from Z-tables or standard normal distribution) is approximately -1.28. The test statistic formula under the alternative hypothesis is $Z = \frac{\bar{X} - \mu_0}{(\sigma/\sqrt{n})}$, where $\bar{X}$ is the sample mean, $\mu_0$ is the hypothesized population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the values ($\mu_0 = 55$, $\sigma = 15$, $n = 49$, and $\mu = 50$), we find the Z value corresponding to $\mu = 50$. The calculation involves finding the Z score for $\mu = 50$ and then finding the area to the left of this Z score (since it's a left-tailed test), which gives us the power. The correct calculation leads to a power of approximately .8531, indicating a high probability of correctly rejecting the null hypothesis when the true mean is 50.

It Is Desired to Test H0:μ=55H _ { 0 } : \mu = 55H0​:μ=55

It Is Desired to Test $H _ { 0 } : \mu = 55$