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The power function for the one-sample \(t\) -test is discussed. (a) Plot the power function for the following setup: \(X\) has a \(N\left(\mu, \sigma^{2}\right)\) distribution; \(H_{0}: \mu=50\) versus \(H_{1}: \mu \neq 50 ; \alpha=0.05 ; n=25 ;\) and \(\sigma=10\). (b) Overlay the power curve in (a) with that for \(\alpha=0.01\). Comment. (c) Overlay the power curve in (a) with that for \(n=35\). Comment. (d) Determine the smallest value of \(n\) so the power exceeds \(0.80\) to detect \(\mu=53\). Hint: Modify the R function tpowerg.R so it returns the power for a specified alternative.

Short Answer

Expert verified
Part (a): Power function plotted with given parameters. Part (b): Power increases with an increase in significance level. The \(\alpha = 0.01\) curve lies above the \(\alpha = 0.05\) curve. Part (c): Power increases with an increase in sample size. The \(n = 35\) curve lies above the \(n = 25\) curve. Part (d): The smallest \(n\) for power > 0.80 for \(\mu = 53\) is obtained by iterating over possible values of \(n\) until the power is greater than 0.80.

Step by step solution

01

Plot power function

Use the given parameters to create a function for the power of the test in R. The function can be plotted using the plot function in R - plot(x, power_fun(x), type='l'), where x is a range of values around the null hypothesis mean. The y-coordinate (power) for each x (alternative mean) can be obtained by using the function pt(), which gives the probability that a t-distributed random variable is less than x.
02

Overlay the curve for different significance level

Change the significance level (\(\alpha\)) to 0.01 in the function, and plot the new power function on the same plot. Compare the two curves to comment on the effect of the significance level on the power of the test. Generally, increasing the significance level would increase the power of the test.
03

Overlay the curve for different sample sizes

Next, change the sample size (\(n\)) to 35 in the function, and plot the new power function on the same plot. Compare the three curves to comment on how the sample size affects the power of the test. Generally, increasing the sample size will increase the power of the test.
04

Determine the smallest sample size for required power

To find the smallest sample size (\(n\)) that gives a power of more than 0.80 to detect a mean of 53, use a while loop in R to increment \(n\) until the function returns a power value of more than 0.80. Note that the 'tpowerg.R' function provided in the exercise might need to be manipulated to be used for this purpose.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{10}\) be a random sample of size 10 from a normal distribution \(N\left(0, \sigma^{2}\right) .\) Find a best critical region of size \(\alpha=0.05\) for testing \(H_{0}: \sigma^{2}=1\) against \(H_{1}: \sigma^{2}=2 .\) Is this a best critical region of size \(0.05\) for testing \(H_{0}: \sigma^{2}=1\) against \(H_{1}: \sigma^{2}=4\) ? Against \(H_{1}: \sigma^{2}=\sigma_{1}^{2}>1 ?\)

Consider a normal distribution of the form \(N(\theta, 4)\). The simple hypothesis \(H_{0}: \theta=0\) is rejected, and the alternative composite hypothesis \(H_{1}: \theta>0\) is accepted if and only if the observed mean \(\bar{x}\) of a random sample of size 25 is greater than or equal to \(\frac{2}{5}\). Find the power function \(\gamma(\theta), 0 \leq \theta\), of this test.

Let \(X_{1}, X_{2}, \ldots, X_{10}\) be a random sample of size 10 from a Poisson distribution with parameter \(\theta\). Let \(L(\theta)\) be the joint pdf of \(X_{1}, X_{2}, \ldots, X_{10}\). The problem is to test \(H_{0}: \theta=\frac{1}{2}\) against \(H_{1}: \theta=1\). (a) Show that \(L\left(\frac{1}{2}\right) / L(1) \leq k\) is equivalent to \(y=\sum_{1}^{n} x_{i} \geq c\). (b) In order to make \(\alpha=0.05\), show that \(H_{0}\) is rejected if \(y>9\) and, if \(y=9\), reject \(H_{0}\) with probability \(\frac{1}{2}\) (using some auxiliary random experiment). (c) If the loss function is such that \(\mathcal{L}\left(\frac{1}{2}, \frac{1}{2}\right)=\mathcal{L}(1,1)=0\) and \(\mathcal{L}\left(\frac{1}{2}, 1\right)=1\) and \(\mathcal{L}\left(1, \frac{1}{2}\right)=2\), show that the minimax procedure is to reject \(H_{0}\) if \(y>6\) and, if \(y=6\), reject \(H_{0}\) with probability \(0.08\) (using some auxiliary random experiment).

Let \(X\) be \(N(0, \theta)\) and, in the notation of this section, let \(\theta^{\prime}=4, \theta^{\prime \prime}=9\), \(\alpha_{a}=0.05\), and \(\beta_{a}=0.10 .\) Show that the sequential probability ratio test can be based upon the statistic \(\sum_{1}^{n} X_{i}^{2} .\) Determine \(c_{0}(n)\) and \(c_{1}(n)\).

Let \(X_{1}, X_{2}, \ldots, X_{10}\) be a random sample from a distribution that is \(N\left(\theta_{1}, \theta_{2}\right)\). Find a best test of the simple hypothesis \(H_{0}: \theta_{1}=\theta_{1}^{\prime}=0, \theta_{2}=\theta_{2}^{\prime}=1\) against the alternative simple hypothesis \(H_{1}: \theta_{1}=\theta_{1}^{\prime \prime}=1, \theta_{2}=\theta_{2}^{\prime \prime}=4\).

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