Find value of the test statistic given set of values and alpha

Question

"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://i.stack.imgur.com/0Vw9p.jpg

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.

Answer 1

Entering your data into R:

x = c(1008,  812, 1117, 1323, 1308, 1415,
       831, 1021, 1287,  851, 930,   740,
       699,  872 , 913,  944,  987,  954, 
      1695,  995, 1003, 994)

The summary statistics required for a t test are as follows:

length(x); mean(x); sd(x)
[1] 22         # sample size
[1] 1031.773   # sample mean
[1] 239.7864   # sample standard deviation

The sample mean $\bar X = 1031.773$ is smaller than the hypothetical population mean $\mu_0 = 1135.$ The question is whether the difference between $\bar X$ and $\mu_0$ is sufficiently large to say that the difference is significantly different at the 5% level.

In R, a one-sample, two-sided t test of $H_0: \mu = 1135$ against $H_a: \mu \ne 1135.$ gives the output shown below. The P-value $0.05641$ is larger than 5%, so you cannot reject $H_0$ at the 5% level of significance. [Other software gives similar output. Or you can do the computation by hand and use printed tables of Student's t distribution to decide whether the reject $H_0.$]

You should make sure you understand how the t statistic is computed and why there are DF = 21 degrees of freedom.

$$T =\frac{\bar X - \mu_0}{S/\sqrt{n}} = \frac{1031.773=1135}{239.7864 /\sqrt{22}} = -2.0192.$$ If $H_0$ is true then $T \sim \mathsf{T}(\nu = n-1 = 21),$ Student's t distribution with DF = 21. Also, be sure you know how you can use P-values to decide whether to reject the null hypothesis.

t.test(x, mu=1135)


        One Sample t-test

data:  x
t = -2.0192, df = 21, p-value = 0.05641
alternative hypothesis: true mean is not equal to 1135
95 percent confidence interval:
  925.4574 1138.0881
sample estimates:
mean of x 
 1031.773 

Also, make sure you know how you can use printed tables to decide whether the reject: On line 21 of the t table you will find that the critical value $c = 2.080$ cuts probability 0.025 from the upper tail of Student's t distribution with 21 degrees of freedom. Thus you reject $H_0$ if $|T| > 2.080.$

qt(.975, 21)
[1] 2.079614

2*pt(-2.0192, 21)
[1] 0.0564132     #  P-value of 2-sided test

Below is a plot of the density function of Student's t distribution with DF = 21. The solid vertical black line show the observed value of the t statistic. The P-value of the test is the sum of the areas in both tails of the distribution outside of the vertical black lines. The vertical red (dashed) lines show the critical values $\pm c = \pm 2.080.$

Answer 2

$\bar X =1031.773,\, \mu=1135,\, S=239.7864,\, n=22.$

$$\frac{\bar X - \mu}{S/\sqrt {n}}=-2.019209$$

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.