91影视

We are given a weight constraint for a substance that its weight should not exceed \(\mu_0 = 5\) kg. We are to test the hypothesis using a control sample \(Y_1, \ldots, Y_n\). The null hypothesis (\(H_0\)) is that the weight constraint is met, and the alternative hypothesis (\(H_1\)) is that the weight constraint is not met. In mathematical terms, these hypotheses are: \(H_0 : \mu = \mu_0 = 5 \text{ kg}\) \(H_1 : \mu < \mu_0\)

We want to find the sample size n such that the type 1 error (rejecting \(H_0\) when it is true) is at most \(5\%\), and the type 2 error (not rejecting \(H_0\) when it is false) is at most \(10\%\) when \(\mu=4.17\). Let's use the standard normal distribution (Z) to represent the normal distribution of the sample mean with a mean of 0 and a standard deviation of 1. We assume that the weights \(Y_i\) are independently and identically distributed as \(N(\mu ; 4)\). We are given the error requirements for the type 1 error, which will be represented by \(\alpha = 0.05\) (\(5\%\)). To find the required sample size n, we use the formula: \(Z = \frac{\bar{Y}_n - \mu_0}{\frac{\sqrt{4}}{\sqrt{n}}}\) where \(\bar{Y}_n\) is the average weight of the sample. Now we will find the corresponding Z value using the type 1 error and the given distribution: \(Z_{\alpha} = -1.645\) Here, we use the negative sign as we are concerned with the lower tail of the distribution based on our alternative hypothesis. The type 2 error (\(\beta\)) is given as \(10\%\), and we are considering the case when \(\mu = 4.17\). We'll denote this value as \(\mu_1\). So, we can write: \(Z_{\beta} = \frac{\bar{Y}_n - \mu_1}{\frac{\sqrt{4}}{\sqrt{n}}}\) For a type 2 error of \(10\%\), the Z value is: \(Z_{\beta} = -1.282\) Now, let's express \(Z_\alpha\) and \(Z_\beta\) in terms of n: \(-1.645 = \frac{\bar{Y}_n - 5}{\frac{2}{\sqrt{n}}}\) \(-1.282 = \frac{\bar{Y}_n - 4.17}{\frac{2}{\sqrt{n}}}\) Subtracting the second equation from the first, we get: \(0.363 = \frac{0.83}{\frac{2}{\sqrt{n}}}\) Solving for n: \(n = \left(\frac{2 \cdot 0.83}{0.363}\right)^2 = 12.97\) Since n cannot be a fraction, we round up: \(n = 13\)

The power function represents the probability of rejecting the null hypothesis for different values of the true parameter. In our case, it is the probability of rejecting \(H_0\) for different values of \(\mu\). We can express the power function as: \(power(\mu) = P\left( \frac{\bar{Y} - \mu_0}{\frac{\sqrt{4}}{\sqrt{n}}} \leq Z_\alpha \right)\) Since we've found n = 13 and \(Z_\alpha = -1.645\), we can use these values along with varying values of \(\mu\) to plot the power function. A visual representation of the power function would show the probability of rejecting \(H_0\) for several values of \(\mu\).