The following 5 questions are based on this information. A sample of 25 cereal b
ID: 3217396 • Letter: T
Question
The following 5 questions are based on this information.
A sample of 25 cereal boxes of Granola Crunch, a generic brand of cereal, yields a mean (X¯) weight of 1.02 pounds. The goal is to construct a 95% confidence interval for the mean weight (µ) of all cereal boxes of Granola Crunch.
Assume that the weight of cereal boxes is normally distributed with a population standard deviation () of 0.03 pounds.
X¯
Select one:
a. 0.006
b. 0.002
c. 0.03
d. 1.02
The critical value (CV) used for a 95% interval estimate is
Select one:
a. 1.96
b. 1.64
c. 0.05
d. 0.025
The 95% confidence interval estimate of µ is
Select one:
a. 1.02 ± 0.012
b. 1.02 ± 0.098
c. 1.02 ± 0.128
d. 1.02 ± 0.059
Suppose you think that average weight of a cereal box of Granola Crunch is 0.9 pounds. In light of the sample evidence and at the 5% level of significance,
Select one:
a. Your claim is not statistically justified
b. Your claim is statistically justified
If we increase the confidence level (1-) from 0.95 to 0.99, the margin of error (ME) of the confidence interval estimate will
Select one:
a. increase
b. stays the same
c. decrease
d. be zero
Please explain in detail step by step for all parts
Explanation / Answer
Solution:
1. a) 0.006
2. a) 1.96
The critical value (CV) used for a 95% interval estimate is 1.96
3. a) 1.02 ± 0.012
ME = CV*SE 0.01176
4. a) Your claim is not statistically justified
When the average weight of a cereal box of Granola Crunch is 0.9 pounds. In light of the sample evidence and at the 5% level of significance, then the claim is not statistically justified.
5. a) increase
If we increase the confidence level (1-) from 0.95 to 0.99, the margin of error (ME) of the confidence interval estimate will increase.