Homework Answers
(3).
Answer:
Given that,
Deon wants to advertise how many chocolate chips are in each Big
Chip cookie at his bakery.
He randomly selects a sample of 50 cookies and finds that the
number of chocolate chips per cookie in the sample has a mean of
15.7 and a standard deviation of 1.4.
What is the 90% confidence interval for the number of chocolate chips per cookie for Big Chip cookies:
Assume the data is from a normally distributed.
The provided sample mean is 
=15.7 and the standard deviation is s=1.4.
The size of the sample is n=50 and the required confidence interval is 90%.
The number of degrees of freedom are df=n-1=50-1=49, and the
level of significance level is 
=0.1.
Based on the provided information, the critical t-value for
=0.1 and df=49 degrees of freedom is tc=1.677.
The 90% confidence interval for the population mean 
is computed using the following expression,
Therefore, based on the information provided, the 90% confidence
interval for the population mean
is,
=[15.7-0.3713, 15.7+0.3713]
=[15.3287, 16.0713]
=[15.329, 16.071]
Note:
** As per HOMEWORKLIB RULES we should solve only the first
question. So I have done it.
For the other question please post differently mentioning your
requirement
Add Answer to:
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