Part (a)
This is a one-tailed test because the alternative hypothesis H1 has a "greater than" symbol.
If it had a "not equal" symbol, then we'd be dealing with a two-tailed test.
Note: the null hypothesis always has an equal sign in it, either in the form of a pure equal sign or a "less than or equal to" or "greater than or equal to" sign.
In this case, the null should read
========================================================================
Part (b)
alpha = 0.02 is the significance level
We're conducting a right-tailed test, so we need to find the value of k such that P(Z > k) = 0.02
We use the standard normal Z distribution because the sample size n is greater than 30 and because we know the population standard deviation sigma.
Use your TI83 or TI84 calculator to compute invNorm(0.02) and you should get roughly -2.053748911
(the invNorm function is found by hitting the blue key labeled "2ND", then hitting the VARS key. This brings up the probability distribution menu)
Change this to a positive number so that we're on the right hand side of the mean mu
So we have 2.053748911 which rounds to 2.0537
This means P(Z > 2.0537) = 0.02 approximately
Decision rule: Reject H0 if the test statistic is larger than z = 2.0537 (otherwise we fail to reject the null H0).
side note: There are other free alternatives if you do not have a TI calculator
For instance, this website
http://onlinestatbook.com/2/calculators/inverse_normal_dist.html
calculates the critical Z value needed. Type 0.02 into the "area" box, 0 for the mean, and 1 for the standard deviation. Then click the "above" radio button. Finally hit "calculate" and you should see 2.054 appear in the box next to the "above" radio button you clicked earlier.
========================================================================
Part (c)
mu = 10 is the hypothesized population mean
xbar = 12 is the sample mean
sigma = 3 is the population standard deviation
n = 36 is the sample size
Let's compute the test statistic
z = (xbar - mu)/(sigma/sqrt(n))
z = (12-10)/(3/sqrt(36))
z = (12-10)/(3/6)
z = (2)/(0.5)
z = 4
Answer: The test statistic is z = 4
========================================================================
Part (d)
Decision: Reject H0
We reject the null because the test statistic z = 4 is larger than the z critical value z = 2.0537
========================================================================
Part (e)
Once again, use your calculator. This time you will use the normalcdf function and you will type in normalcdf(4,99) which effectively computes the area under the standard normal Z curve from z = 4 onward to z = 99 (which is so far away from the mean z = 0 that it's effectively infinity). This will allow us to compute P(Z > 4).
This is what your calculator should look like
(the normalcdf function is found in the same menu as the invNorm function)
The result 3.168603459E-5 means we have 3.168603459 * 10^(-5) which converts to 0.0000316860346 and then that rounds to 0.0000317
This is a very small P value which helps confirm our decision to reject the null.
Recall that we reject the null whenever the P value is smaller than alpha.
Answer: The P value is approximately 0.0000317
One informal way to interpret the P value is to think of it as the chances of getting the null correct. It's not entirely accurate but its close enough to work. So you can think of it as a probability of 0.0000317 that you get the null correct. This is a very small chance and you're much better off accepting the alternative hypothesis instead.
Because we rejected the null and accepted the alternative hypothesis, this means the population mean (mu) is likely larger than 10.