In: Statistics and Probability
Jerry is a district manager for Home Depot and he recently implemented a new safety program in the 8 stores he manages. He would like to know if this new program has resulted in fewer accidents than other Home Depot stores. Jerry knows that the average number of accidents at any Home Depot is 4.2 a month with a standard deviation of 1.1. Jerry’s has the following accident data for his 8 stores. Use a single-sample z-test to determine the outcome (alpha = .05, one-tailed).
Store 1: 2 accidents
Store 2: 1 accidents
Store 3: 4 accidents
Store 4: 3 accidents
Store 5: 5 accidents
Store 6: 1 accidents
Store 7: 0 accidents
Store 8: 2 accidents
provide the following information:
Null Hypothesis in sentence form :
Alternative Hypothesis in sentence form :
Critical Value(s) :
Calculations : Note: the more detail you provide, the more partial credit that I can give you if you make a mistake.
Outcome (determination of significance or not, and what this reflects in everyday language,
Important notations
mean number of accicidents for all the stores.
standard deviation of the number of accicidents for all the
stores.
sample mean for the 8 sample stores
s sample standard deviation for the 8 sample stores
n sample size.
level of significance
Now we have the information that
So the null hypothesis,
Or the null hypothesis states that there is no effect of his new program. Accident rate remains the same.
The alternative hypothesis
Alternative hypothesis says this new program has resulted in fewer accidents than other Home Depot stores.
the test statistic is
Now i have performed the test in MINITAB software:
Steps required:
So the minitab output is
So,
From minitab output,
the test statitsic z0=-5.01
the crittcal value is
As,
we reject the null hypothesis
.
And accept the alternative hypothesis
So can conclude that the test is significant and,
"this new program has resulted in fewer accidents than other Home Depot stores".
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