Question 1193127: Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 34 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 9.1 and a standard deviation of 3.5. What is the 80% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Enter your answers accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
The sample Karen selected has these properties- xbar = 9.1 = sample mean
- s = 3.5 = sample standard deviation
- n = 34 is the sample size
xbar estimates mu (the population mean).
s estimates sigma (the population standard deviation).
Because n > 30, this means we can use the Z distribution even though we don't know sigma.
At 80% confidence, we have z = 1.282 as the approximate z critical value.
Use a Z table to determine this.
This value means P(-1.282 < Z < 1.282) = 0.80 approximately.
E = margin of error
E = z*s/sqrt(n)
E = 1.282*3.5/sqrt(34)
E = 0.76951415153443
E = 0.769514 approximately
L = lower bound of the confidence interval
L = xbar - E
L = 9.1 - 0.769514
L = 8.330486
L = 8.3
U = upper bound of the confidence interval
U = xbar + E
U = 9.1 + 0.769514
U = 9.869514
U = 9.9
When written in the format (L, U), the confidence interval is roughly (8.3, 9.9)
Alternatively, you can opt for the L < mu < U format to write 8.3 < mu < 9.9
I like the second notation better because it involves mu to tell the reader mu is likely somewhere between 8.3 and 9.9 (we are 80% confident of this).
However, many textbooks and scientific journals opt for the condensed format mentioned in the first option.
This is likely the format your teacher will want. Of course, if your teacher wants otherwise, then be sure to follow such instructions.
Answer: (8.3, 9.9)
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