Find value of test statistic given significance and unknown mean amount

Question

"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.

Answer 1

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.