Question

You measure 46 watermelons' weights, and find they have a mean weight of 60 ounces. Assume the population standard deviation is 8.5 ounces. Based on this, construct a 99% confidence interval for the true population mean watermelon weight. Give your answers as decimals, to two places.

Answer 1

A 99% confidence for the true population mean watermelon weight is [56.77 ounces, 63.23 ounces] .

Explanation:

We are given that you measure 46 watermelons' weights, and find they have a mean weight of 60 ounces.

Assume the population standard deviation is 8.5 ounces.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. =
`\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{} n} }` ~ N(0,1)

where,
`\bar X` = sample mean weight = 60 ounces

`\sigma` = population standard deviation = 8.5 ounces

n = sample of watermelons = 46

`\mu` = population mean watermelon weight

Here for constructing a 99% confidence interval we have used a One-sample z-test statistics because we know about the population standard deviation.

So, 99% confidence interval for the population mean,
`\mu` is ;

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.58 & 2.58}

P(-2.58 <
`\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{} n} }` < 2.58) = 0.99

P(
`-2.58 * {\frac{\sigma}{\sqrt{} n} }` <
`{\bar X-\mu}` <
`2.58 * {\frac{\sigma}{\sqrt{} n} }` ) = 0.99

P(
`\bar X-2.58 * {\frac{\sigma}{\sqrt{} n} }` <
`\mu` <
`\bar X+2.58 * {\frac{\sigma}{\sqrt{} n} }` ) = 0.99

99% confidence interval for
`\mu` = [
`\bar X-2.58 * {\frac{\sigma}{\sqrt{} n} }` ,
`\bar X+2.58 * {\frac{\sigma}{\sqrt{} n} }` ]

= [
`60-2.58 * {(8.5)/(√(46) ) }` ,
`60+2.58 * {(8.5)/(√(46) ) }` ]

= [56.77 ounces, 63.23 ounces]

Therefore, a 99% confidence for the true population mean watermelon weight is [56.77 ounces, 63.23 ounces] .