Suppose you have a sample of size 42 with a mean of 30 and a population standard deviation of 7.4. What is the maximal margin of error associated with a 95% confidence interval for the true population mean? As in the reading, in your calculations, use z = 2. Give your answer as a decimal, to two places

Answer:The maximal margin of error associated with a 95% confidence interval for the true population mean is 2.238.Step-by-step explanation:We have given,The sample size n=42The sample mean [tex]\bar{x}=30[/tex]The population standard deviation [tex]\sigma=7.4[/tex]Let [tex]\alpha[/tex] be the level of significance = 0.05Using the z-distribution table,The critical value at 5% level of significance and two tailed z-distribution is [tex]\pm z_{\frac{0.05}{2}}=\pm 1.96[/tex]The value of margin of error is [tex]ME=z_{\alpha/2}(\frac{\sigma}{\sqrt{n}})[/tex][tex]ME=1.96(\frac{7.4}{\sqrt{42}})[/tex][tex]ME=1.96(1.1418)[/tex][tex]ME=2.238[/tex]The maximal margin of error associated with a 95% confidence interval for the true population mean is 2.238.