A random sample of 80 observations from a population with a population mean 198 and a population standard deviation of 15 yielded a sample mean of 190.

a. Construct a hypothesis test with the alternative hypothesis that\(\mu < 198\)at a 1% significance level. Interpret your results.

b. Construct a hypothesis test with the alternative hypothesis that

at a 1% significance level. Interpret your results.

c. State the Type I error you might make in parts a and b.

The random sample size\(n = 64\)has a sample mean\(\bar x = 190\).

The population mean and the standard deviation is 198 and 15, respectively.

The two-sided and one-sided hypothesis testing problems \(\mu = 198\) need to test at a 1% significance level.

The test statistic is:

\(\begin{aligned}z &= \frac{{\bar x - \mu }}{{\frac{\sigma }{{\sqrt n }}}}\\ &= \frac{{190 - 198}}{{\frac{{15}}{{\sqrt {80} }}}}\\ &= \frac{{ - 8}}{{1.6771}}\\ &= - 4.77\end{aligned}\)

Therefore, the test statistic is \(z = - 4.77\).

a.

The p-value for the left-tailed test is:

\(\begin{aligned}p &= P\left( {Z < - 4.77} \right)\\ < 0.0001\end{aligned}\).

The probability is obtained by using the calculators.

Since the p-value is less than 0.01, reject the null hypothesis\(\mu = 198\).

Therefore, there is sufficient evidence to support \(\mu < 198\).

b.

The p-value for the two-tailed test is:

\(\begin{aligned}p & = 2 \times P\left( {Z < - 4.77} \right)\\ < 2 \times 0.0001\\ < 0.001\end{aligned}\).

The p-value is smaller than 0.01; therefore, reject the null hypothesis\(\mu = 198\).

Hence, there is enough evidence to support this \(\mu \ne 198\).

c.

The Type I error that would occur in part a is rejecting the population mean\(\mu = 198\); in fact, it is true.

Also, for part b, Type I error that occurs is researcher supports the claim that \(\mu \ne 198\); in fact, \(\mu = 198\).