12. Makayla owns a gardening store. She has a machine that automatically fills bags of mulch. The bags of
mulch should weigh 10 lbs., but Makayla suspects the machine has not been correctly filling the bags to
that weight. She randomly samples 25 bags and weighs them. She calculates a sample mean of 9.65 lbs.
with a sample standard deviation of 0.65 lbs. Perform a significance test with a confidence level of a=0.01 and make a conclusion.
Null Hypothesis: The machine is correctly filling the bags of mulch to an average weight of 10 lbs.
Alternative Hypothesis: The machine is not correctly filling the bags of mulch to an average weight of 10 lbs.
We will use a t-test since the sample size is less than 30 and the population standard deviation is unknown.
t = (x̄ - μ) / (s / √n)
t = (9.65 - 10) / (0.65 / √25)
t = -1.5
Using a t-table with 24 degrees of freedom (25-1), at a significance level of 0.01, the critical t-value is ±2.492.
Since our calculated t-value (-1.5) is not greater than the critical t-value (-2.492) or less than (-2.492), we fail to reject the null hypothesis.
Conclusion: We do not have sufficient evidence to conclude that the machine is not correctly filling the bags of mulch to an average weight of 10 lbs.