Question

Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is reasonable to assume that the population is approximately normal.

21.88 21.76 22.14 21.63 21.81 22.12 21.97 21.57 21.75 21.96 22.20 21.80

Required:
Construct a 90% confidence interval for the mean weight.

Answer 1

The 99% confidence interval for the mean weight is (21.7025, 22.0625)

Given that,

Sample mean
`\bar x` = 21.8825

Sample standard deviation s = 0.200732

Sample size n = 12

Degrees of freedom df = 12 -1 = 11

We have to construct a 99% confidence interval for the mean weight.

Now,

99% = 99/100 = 0.99

Here,

a = 10.99 = 0.01

The formula for the confidence interval is (
`\bar x` E,
`\bar x` + E)

Degrees of freedom formula is df = n - 1

The formula for the confidence interval is (
`\bar x` - E,
`\bar x` + E)

Here population standard deviation is not known so we use t-interval.

To find the t-value Use the Excel function = TINV(a, df)

So, TINV(0.01, 11)

We get,

tc = 3.106

The margin of error is given by

E =
`(t_c * s)/(√(n) )`

E =
`(t_c * 0.200732)/(√(n) )`

E =
`(0.623473)/(3.46410)`

E = 0.17998

Thus,

The Confidence interval is given by

CI=(
`\bar x` - E,
`\bar x` + E)

= (21.8825-0.17998, 21.8825 +0.17998)

= (21.7025, 22.0625)

Answer 2

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

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

P.Q. =
`(\bar X-\mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where,
`\bar X` = sample mean weight =
`(\sum X)/(n)` = 21.88 ounces

s = sample standard deviation =
`\sqrt{(\sum (X-\bar X)^(2) )/(n-1) }` = 0.201 ounces

n = sample of boxes = 12

`\mu` = population mean weight

Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

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

P(-1.796 <
`t_1_1` < 1.796) = 0.90 {As the critical value of t at 11 degrees of

freedom are -1.796 & 1.796 with P = 5%}

P(-1.796 <
`(\bar X-\mu)/((s)/(√(n) ) )` < 1.796) = 0.90

P(
`-1.796 * {(s)/(√(n) ) }` <
`{\bar X-\mu}` <
`1.796 * {(s)/(√(n) ) }` ) = 0.90

P(
`\bar X-1.796 * {(s)/(√(n) ) }` <
`\mu` <
`\bar X+1.796 * {(s)/(√(n) ) }` ) = 0.90

90% confidence interval for
`\mu` = [
`\bar X-1.796 * {(s)/(√(n) ) }` ,
`\bar X+1.796 * {(s)/(√(n) ) }` ]

= [
`21.88-1.796 * {(0.201)/(√(12) ) }` ,
`21.88+1.796 * {(0.201)/(√(12) ) }` ]

= [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].