Question 1192022
claim is that meal preparation time has a mean of 22 hours.
sample size is 28.
sample mean is 17.5
sample standard deviation is 4.3  hours.
since you are using sample standard deviation rather than population standard deviation, the use of the t-score is indicated.
degrees of freedom is sample size minus 1 = 27.
critical t-score at .05 significance level with 27 degrees of freedom is equal to -1.70.
standard error = standard deviation of sample / square root of sample size = 4.3 / sqrt(38) = .8126.
test t-score = (17.5 - 22) / .8126 = -5.54.
this is greater than the critical t-score, indicating that the average preparation time of means is less than 22 hours.
that's the conclusion.