Question

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 (σ) of 0.03 pounds.

Answer 1

Answer:

1.01 < µ < 1.03

Step-by-step explanation:

We want to construct a 95% confidence interval.  Our sample size is 25, so we use a t-value.  The degrees of freedom are 24 (always one less than the sample size).  

The t-value we get is:  2.064

We have:

x = 1.02

σ = 0.03

See the attached photo for the construction of the confidence interval...

Answer 2

95% confidence interval for cereal boxes crunch is : ( 0.896 , 1.1438 )

Given : Mean x' = 1.02 , Standard Deviation s = 0.03 , Confidence Limit = 95% , n = 25

Confidence Internal = x' + t ( s / √n)

t at 95% confidence limit, ie 5% ( 0.05 ) critical value  

t at 0.05/ 2 for two sided = t 0.025 , at degrees of freedom 25 - 1 ie 24 = 2.06390

• Confidence Internal  = 1.02 + 2.06390 ( 0.03 / √25 )

= 1.02 + 0.123834

( 0.896166 , 1.143834 )

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