Find value of test statistic given significance and unknown mean amount "Suppose that in past years the average price per square metre for warehouses in Canada has been 347.46 dollars. A national real estate investor wants to determine whether that figure has changed now. The investor hires a researcher who randomly samples 48 warehouses that are for sale across Canada and finds that the mean price per square foot is 339.80 dollars, with a standard deviation of $13.89. Assume that prices of warehouse area are normally distributed in population. If the researcher uses a 5% level of significance, what statistical conclusion can be reached?"
Attempt:
n=48
Xbar= $339.80
o=$13.89
a=0.05
(xbar-u)/(o/rootn)=3.82=test statistic
Reject null hypothesis
The numerical answer is wrong so what mistake did I make? Whats the correct equation to solve for test equation.
 A: Minitab statistical software will accept summarized data, such as yours, so we can check our work against Minitab output later.
You are doing a t test because the population standard deviation $\sigma$ is unknown; instead you are given the sample standard deviation $S,$ which estimates $\sigma.$
I believe you are testing
$H_0: \mu = 347.40$ against $H_a:\mu < 347.40$ at the 5% level. The critical value $c = -1.6779.$
cuts probability $\alpha = 0.05$ from the lower tail of Student's t distribution
with 47 degrees of freedom. You will reject $H_0$ if the $T$ statistic is smaller (more negative) than $c.$ [You can find the critical value
in a printed table of t distributions or by using software.]
The $T$ statistic is
$$T = \frac{\bar X =\mu_0}{S/\sqrt{n}} =
\frac{339.80 - 347.40}{13.89/\sqrt{48}}= -3.791.$$
Minitab output is shown below:
One-Sample T 

Test of μ = 347.4 vs < 347.4

 N    Mean  StDev  SE Mean  95% Upper Bound      T      P
48  339.80  13.89     2.00           343.16  -3.79  0.000

The P-value is the probability below $-3.79$ in the lower tail of Student's t distribution with DF = 47. You will
reject $H_0$ at the 5% level if the P-value is smaller
than 5%. [Generally, it is not possible to find exact P-values from printed tables. One usually gets P-values from software.]
In R statistical software you can get the P-value as shown below. Minitab rounds this to 0.000.
pt(-3.791, 48)
[1] 0.0002096364

In this problem there are two methods for knowing to
reject $H_0$ at the 5% level:
(a) $T < c;$
(b) P-value $< 0.05 = 5\%.$
Without software, one generally uses method (a); with software, method (b).
You should understand both.
