SOLUTION:
A recent article in a leading magazine claimed that the homemakers spend an
average of 22 hours per week preparing meals. A director of marketing at a foodprocessing compan
Algebra ->
Probability-and-statistics
-> SOLUTION:
A recent article in a leading magazine claimed that the homemakers spend an
average of 22 hours per week preparing meals. A director of marketing at a foodprocessing compan
Log On
Question 1192022:
A recent article in a leading magazine claimed that the homemakers spend an
average of 22 hours per week preparing meals. A director of marketing at a foodprocessing company surveyed that a recent sample of 28 homemakers resulted in
an average time preparation of meals of 17.5 hours per week, with a standard
deviation of 4.3 hours. Can the director conclude that the average preparation of
meals is lesser? Use 0.05 significance level Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! 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.
