4.1. suppose that you are testing the following hypotheses where the variance is known: $h_0:mu = 100$…

Explanation:

Step1: Recall P - value for two - tailed test

For a two - tailed test with null hypothesis $H_0:\mu = 100$ and alternative hypothesis $H_1:\mu\neq100$, the P - value is $2\times(1 - \Phi(|z_0|))$, where $\Phi$ is the cumulative distribution function of the standard normal distribution and $z_0$ is the test statistic.

Step2: Calculate P - value for $z_0 = 2.75$

Using the standard normal table, $\Phi(2.75)=0.9970$. Then $P - value=2\times(1 - 0.9970)=2\times0.0030 = 0.0060$.

Step3: Calculate P - value for $z_0 = 1.86$

From the standard normal table, $\Phi(1.86)=0.9686$. So $P - value=2\times(1 - 0.9686)=2\times0.0314 = 0.0628$.

Step4: Calculate P - value for $z_0=-2.05$

Since the standard normal distribution is symmetric, $\Phi(| - 2.05|)=\Phi(2.05)=0.9800$. Then $P - value=2\times(1 - 0.9800)=2\times0.0200 = 0.0400$.

Step5: Calculate P - value for $z_0=-1.86$

As $\Phi(| - 1.86|)=\Phi(1.86)=0.9686$, $P - value=2\times(1 - 0.9686)=2\times0.0314 = 0.0628$.

Answer:

a. $0.0060$ b. $0.0628$ c. $0.0400$ d. $0.0628$