Question

You measure 50 textbooks' weights, and find they have a mean weight of 37 ounces. Assume the population standard deviation is 5.2 ounces. Based on this, construct a 90% confidence interval for the true population mean textbook weight. Give your answers as decimals, to two places

Answer 1

90% confidence interval for the true population mean textbook weight is [35.79 ounces , 38.21 ounces].

Explanation:

We are given that you measure 50 textbooks' weights, and find they have a mean weight of 37 ounces.

Assume the population standard deviation is 5.2 ounces.

Firstly, the Pivotal quantity for 90% confidence interval for the population mean is given by;

P.Q. =
`(\bar X-\mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\bar X` = sample mean weight = 37 ounces

`\sigma` = population standard deviation = 5.2 ounces

n = sample of textbooks = 50

`\mu` = true population mean textbook weight

Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 90% confidence interval for the population mean,
`\mu` is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5%

level of significance are -1.645 & 1.645}

P(-1.645 <
`(\bar X-\mu)/((\sigma)/(√(n) ) )` < 1.645) = 0.90

P(
`-1.645 * {(\sigma)/(√(n) ) }` <
`{\bar X-\mu}` <
`1.645 * {(\sigma)/(√(n) ) }` ) = 0.90

P(
`\bar X -1.645 * {(\sigma)/(√(n) ) }` <
`\mu` <
`\bar X +1.645 * {(\sigma)/(√(n) ) }` ) = 0.90

90% confidence interval for
`\mu` = [
`\bar X -1.645 * {(\sigma)/(√(n) ) }` ,
`\bar X +1.645 * {(\sigma)/(√(n) ) }` ]

= [
`37-1.645 * {(5.2)/(√(50) ) }` ,
`37+1.645 * {(5.2)/(√(50) ) }` ]

= [35.79 , 38.21]

Therefore, 90% confidence interval for the true population mean textbook weight is [35.79 ounces , 38.21 ounces].