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=42
The 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 = 0.05
Using 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.
