Question

1. A company that produces bread is concerned about the distribution of the amount of sodium in its bread. The company takes a simple random sample of 100 slices of bread and computes the sample mean to be 103 milligrams of sodium per slice. Assume that the population standard deviation is 10 milligrams. a) Construct a 95% confidence interval estimate for the mean sodium level. b) Construct a 99% confidence interval estimate for the mean sodium level.

Answer 1

The 95% confidence interval is
`101.13  <  \mu < 104.87`

The 99% confidence interval is
`100.54  <  \mu < 105.46`

Explanation:

From the question we are told that

The sample size is n = 110

The sample mean is
`\= x = 103 \ mg`

The population standard deviation is
`\sigma = 10 \ mg`

From the question we are told the confidence level is 95% , hence the level of significance is

`\alpha = (100 - 95 ) \%`

=>
`\alpha = 0.05`

Generally from the normal distribution table the critical value of
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  1.96`

Generally the margin of error is mathematically represented as

`E = Z_{(\alpha )/(2) } *  (\sigma )/(√(n) )`

=>
`E = 1.96 *  ( 10 )/(√(110) )`

=>
`E =1.8688`

Generally 95% confidence interval is mathematically represented as

`\= x -E <  \mu <  \=x  +E`

=>
`103 - 1.8688 <  \mu < 103 + 1.8688`

=>
`101.13  <  \mu < 104.87`

Considering question b

From the question we are told the confidence level is 99% , hence the level of significance is

`\alpha = (100 - 99 ) \%`

=>
`\alpha = 0.01`

Generally from the normal distribution table the critical value of
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  2.58`

Generally the margin of error is mathematically represented as

`E = Z_{(\alpha )/(2) } *  (\sigma )/(√(n) )`

=>
`E = 2.58 *  ( 10 )/(√(110) )`

=>
`E =2.4599`

Generally 99% confidence interval is mathematically represented as

`\= x -E <  \mu <  \=x  +E`

=>
`103 - 2.4599 <  \mu <103 + 2.4599`

=>
`100.54  <  \mu < 105.46`