Question

You measure 41 textbooks' weights, and find they have a mean weight of 30 ounces. Assume the population standard deviation is 12.3 ounces. Based on this, construct a 93% confidence interval for the true population mean textbook weight. Round z-values to 2 decimal places; round t-values to 3 decimal places. Round final answers to 2 decimal places.

Answer 1

The 93% confidence interval is
`26.52   <  \mu < 33.48`

Explanation:

From the question we are told that

The sample size is n = 41

The mean is
`\= x = 30`

The standard deviation is
`\sigma = 12.3`

From the question we are told the confidence level is 93% , hence the level of significance is

`\alpha = (100 - 93 ) \%`

=>
`\alpha = 0.07`

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

`Z_{(\alpha )/(2) } =  1.81`

Generally the margin of error is mathematically represented as

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

=>
`E =  1.812  *  (12.3)/(√(41) )`

=>
`E =  3.48`

Generally 93% confidence interval is mathematically represented as

`\= x -E <  \mu <  \=x  +E`

=>
`30  -3.48  <  \mu < 30  + 3.48`

=>
`26.52   <  \mu < 33.48`