Question 1183057
the sample mean is 22.625 and the sample standard deviation is 4.9339.
the sample size is 40.
test mean is 21.21 and test is whether the sample mean is greater than that.


test is whether sample mean is greater than test mean.


standard error = sample standard deviation / sqrt(sample size) = 4.9339 / sqrt(40) = .780118 rounded to 6 decimal places.


t-score = (x - m) / s
x is the samplemean.
m is the test mean
s is the stnadard error


t = (22.625 - 21.21) / .780118 = 1.814 rounded to 3 decimal places.


critical p-value is .05
test is one tailed, therefore critical p-value is .05 on the high end, because the test is whether the sample mean is greater than the test mean.


h0 = 21.21
ha = 22.625


degrees of freedom = 40 minus 1 = 39


p-value for area to the right of t-value of 1.814 with 39 degrees of freedom is .0387 rounded to 4 decimal places.


since this is less than .05, results are significant and alternate hypothesis is accepted.


here are the results from the t-test calculator at <a href = "https://www.socscistatistics.com/pvalues/tdistribution.aspx" target = "_blank">https://www.socscistatistics.com/pvalues/tdistribution.aspx</a>


<img src = "http://theo.x10hosting.com/2021/071804.jpg" >


test statistic is t-value of 1.814 rounded to 3 decimal places.
test p-value is .039 rounded to 3 decimal places.


any questions, send me a message.