to gets descriptive statistics on the data, i used online calculator at http://www.alcula.com/calculators/statistics/dispersion/#gsc.tab=0
here are the results.
results from that calculaor that i needed are:
sample size is 10.
mean of sample = 18
standard deviation of sample = 3.7118429085533 which i rounded to 3.71184.
since we are interested in the mena of the sample, we need to use the standard error.
standard error = standard deviation of sample / sqrt(sample size) = 3.71184 / sqrt(10) = 1.173578161 rounded to 1.173578.
t-score = (18 - 20 / 1.173578 = -1.704 rounded to 3 decimal places.
t-score formula is t = (x - m) / s
t is the t-score
x is the sample mean
m is the claimed population mean.
s is the standard error.
since the sample size is 10, the degrees of freedom is 1 less than that = 9.
probability of getting a sample mean less than 18, when the population mean is 20, is equal to .06128792431 which i rounded to .06129.
i used my ti-84 polus calculator to get this result.
i also used the one sample t-test calculator at https://www.statskingdom.com/130MeanT1.html#google_vignette
here are the results from that calculator.
note that the calculator states that the test p-value = .1226.
my test p-value was equal to .0613.
the difference is that i am looking at the p-value on the left side of the confidence interval and the calculator is stating the p-value for both sides of the confidence interval.
the real p-value of the test is .1226 as the calculator stated, because it encompasses both tails of the two-tail confidence interval.
if i had stated up front that the criticl p-value was .05, then a two-tail alpha would be .025 on the left and .025 on the right.
.1226 is greater than .05 while .0613 is greater than .025, both indicatging that the test p-value is greater than the critical p-value which concludes the results of the test are not significant enough to state that the true mean is not 20, but something different.