as i can not upload a file so cookies.txt file look like this 49 49.5 46.4 47.2
ID: 3267900 • Letter: A
Question
as i can not upload a file so cookies.txt file look like this
49
49.5
46.4
47.2
48.2
51.6
50.5
47.4
50.3
46.6
47.6
53.1
45.3
48
49.2
49.5
49.2
49.8
50.1
47.8
49.9
48.8
46.3
47.3
42.4
as i can not upload a file so cookies.txt file look like this
49
49.5
46.4
47.2
48.2
51.6
50.5
47.4
50.3
46.6
47.6
53.1
45.3
48
49.2
49.5
49.2
49.8
50.1
47.8
49.9
48.8
46.3
47.3
42.4
2. A baker is concerned that his chocolate chip cookies weigh less, on average, than their advertised weight of 50 grams each, and that he will lose customers as a result. Assume that cookies, weights (in grams) are independent and distributed as N(, *) a) (5 marks) The baker guesses from historical data that is at most 3 . He considers a mean weight of less than 48 grams to be problematic. What is the minimum number of cookies that he will need to weigh in order to be able to detectExplanation / Answer
a. It is related to minimum sample size selection problem
n=(z*sigma/Error)^2
n=(1.96*3/2)^2 = 8.6= 7 cookies
error=50-48=2
b. We will use z-test since n=30
H0: mean weight of cookies >=50
Ha: mean weight of cookies <50 (left-tailed test)
test stat z= (x-Mu)/(sigma/sqrt(n))
p-value<5%, we reject H0 and conclude that mean cookies weighs less than 50 grms.
Formula:
c. Type-I error: rejecting H0 when it's true: concluding that mean weight is <50, but actually it's not
Type-II error: Not rejecting H0 when it's false: concluding that mean weight is >=50, but actually it's not
In this case Type-II is more dangerous since cookies having less weight will spoil baker's image.
49 mean= 48.44 49.5 SD= 2.176962716 46.4 SE= 0.397457195 47.2 Point estimate= -1.56 48.2 test stat z= -3.92495096 51.6 p-value= 4.33737E-05 50.5 47.4 50.3 46.6 47.6 53.1 45.3 48 49.2 49.5 49.2 49.8 50.1 47.8 49.9 48.8 46.3 47.3 42.4Related Questions
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