Explain please.. A sample of 25 cereal boxes of Granola Crunch, a generic brand

Question

Explain please..

A sample of 25 cereal boxes of Granola Crunch, a generic brand of cereal, yields a mean (X¯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¯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

Explanation / Answer

n = 25 , s = 0.03 , mean = 1.02

Standard error = s / sqrt(n)

= 0.03 / sqrt(25)

= 0.006

Answer is option A)

Step 1: Subtract the confidence level from 100% to find the level: 100% – 95% = 5%.

Step 2: Convert Step 1 to a decimal: 5% = 0.05.

Step 3: Divide Step 2 by 2 (this is called “/2”).
0.05 = 0.025. This is the area in each tail.

Step 4: Subtract Step 3 from 1 (because we want the area in the middle, not the area in the tail):
1 – 0.025 = .975.

Step 5: Look up the area from Step in the z-table. The area is at z=1.96. This is your critical value for a confidence level of 95%.

CI = mean + / -z * (s/sqrt(n))

= 1.02 + /- 1.96 * ( 0.03 / sqrt(25))

= 1.02 + / - 0.012

Answer is option A)

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,   

a. Your claim is not 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

a. increase

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