Answer:
There is 8% (P=0.08) that Frances concludes that the new equipment increases the average daily jewelry production when in fact the new equipment has no effect.
Step-by-step explanation:
We have one-sample z-test with a significance level of 0.08 and a power ot the test of 0.85.
In this test, the null hypothesis will state that the new equipment has the same productivity of the older equipment. The alternative hypothesis is that there is a significative improvement from the use of new equipment.
The probability that Frances concludes that the new equipment increases the average daily jewelry production when in fact the new equipment has no effect is equal to the probability of making a Type I error (rejecting a true null hypothesis).
The probability of making a Type I error is defined by the level of significance, and in this test this value is α=0.08.
Then, there is 8% that Frances concludes that the new equipment increases the average daily jewelry production when in fact the new equipment has no effect.
Question Completion:
How are the percentages distributed? Is the distribution skewed? Are there any gaps? (Select all that apply.)
Answer:
1. The percentages are concentrated from 20% to 60%.
2. These data are strongly skewed left.
3. There are no gaps in the data.
Step-by-step explanation:
1. Data
Percentage loss of wetlands per state
46 37 36 42 81 20 73 59 35 50
87 52 24 27 38 56 39 74 56 31
27 91 46 9 54 52 30 33 28 35
35 23 90 72 85 42 59 50 49
48 38 60 46 87 50 89 49 67
2. Re-arrangement of
Percentage loss of wetlands per state (in ascending order)
9 20 23 24 27 27 28 30
31 33 35 35 35 36 37 38
38 39 42 42 46 46 46 48
49 49 50 50 50 52 52 54
56 56 59 59 60 67 72 73
74 81 85 87 87 89 90 91
1/2 of the cake was eaten
1+2=3. 3/6=1/2
All the slices are the same size
you would find the mean of all of them and it would be 7.6
Just times 16 and 5 ft. together to get your answer which is 80
