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Question: (A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each is weighed. The sample mean was found to be 410 grams and the sample standard deviation was 40 grams. Find the 90% confidence interval for the mean weight of shipped homemade candies. (Round your final answers to the nearest hundredth)

Question:

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each is weighed. The sample mean was found to be 410 grams and the sample standard deviation was 40 grams. Find the 90% confidence interval for the mean weight of shipped homemade candies. (Round your final answers to the nearest hundredth)
(B) When 500 college students are randomly selected and surveyed; it is found that 155 own a car. Find a 90% confidence interval for the true proportion of all college students who own a car.

(Round your final answers to the nearest hundredth)

 

(C) Interpret the results (the interval) you got in (A) and (B)

 

Answer:

Solution(A)
Given in the question
Number of random sample (n) = 16
Sample mean (Xbar) = 410
Sample standard deviation (S) = 40
We need to calculate 90% confidence interval for the mean weight of shipped homemade candies which can be calculated as
Xbar +/- t\alpha/2 * S/sqrt(n)
Here confidence level = 0.90
Level of significance (\alpha) = 1 – 0.90 = 0.10
\alpha/2 = 0.10/2 = 0.05
degree of freedom (df) = n-1 = 16-1 = 15
from t table we found t\alpha/2 = 1.753
So 90% confidence interval is
410 +/- 1.753 * 40/sqrt(16)
410 +/- 17.53
392.47 < \mu < 427.53
So 90% confidence interval for the mean weight of shipped homemade candies is between 392.47 and 427.53

Solution(B)
Given in the question
Number of sample (n) = 500
Number of students have own car (X) = 155
Sample proportion (p’) = X/n = 155/500 = 0.31
Here we need to calculate 90% confidence interval for the true proportion of all college students who own a car which can be calculated as
p’ +/- Z\alpha/2 * sqrt(p'(1-p’)/n)
Here confidence level = 0.90
Level of significance (\alpha) = 0.10
\alpha/2 = 0.05
from Z table we found Z\alpha/2 = 1.645
So 90% confidence interval for the true proportion of all college students who own a car is
0.31 +/- 1.645*sqrt(0.31*(1-0.31)/500)
0.31 +/- 0.034
0.276 < p < 0.344
So 90% confidence interval for the true proportion of all college students who own a car is between 0.28 and 0.34

Solution(c)
In part A, we are 90% confident that true population mean weight of shipped homemade candies is between 392.47 grams and 427.53 grams.
In part B, we are 90% confident that true population proportion of all college students who own a car is between 0.28 and 0.34

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