Find value of the test statistic given set of values and alpha "Based on population figures and other general information on the Canadian population, suppose it has been estimated that, on average, a family of four in Canada spends about $1,135 annually on dental expenditures. Suppose further that a regional dental association wants to test to determine if this figure is accurate for their area of the country. To test this, 22 families of four are randomly selected from the population in that area of the country and a log is kept of the family’s dental expenditures for one year. The resulting data are given below. Assuming that dental expenditures are normally distributed in the population, use the data and an alpha of 0.05 to test the dental association’s hypothesis."
1,008   812 1,117   1,323   1,308   1,415
831 1,021   1,287   851 930 740
699 872 913 944 987 954
1,695   995 1,003   994 

Attempt:
Did https://imgur.com/a/f6tJ8vD
Then use (xbar-u)/(o/rootn)
n=22
o=245.6477
xbar=1135
u=60342.79
(1135-60342.79)/(245.6477/root22)
=-1130.52
Fail to reject null
The numerical answer is wrong so what mistake did I make? Whats the correct equation to solve for test equation.

$\bar X =1031.773,\,
\mu=1135,\, 
S=239.7864,\,
n=22.$
$$\frac{\bar X - \mu}{S/\sqrt {n}}=-2.019209$$
Assume the hypotheses
$$H_0: \mu = 1135\\
H_1: \mu\ne1135$$
By statistical software you can get a p-value of $2\times Pr(t_{21}<-2.02)=0.056$. Since it's greater than $\alpha=.05$, you wouldn't reject the null. So the figure for the average annual dental expenditure in Canada for a family of four is valid for the area of the country examined by the regional dental association.
Instead if you were to do a one-sided alternative hypothesis $H_1: \mu < 1135$ you would not multiply by $2$ and end up rejecting the null.
