Question

In a metal fabrication​ process, metal rods are produced to a specified target length of 15 feet. Suppose that the lengths are normally distributed. A quality control specialist collects a random sample of 16 rods and finds the sample mean length to be 14.8 feet and a standard deviation of 0.65 feet. What is the​ 95% confidence interval for the true mean length of rods produced by this​ process?

Answer 1

95% confidence interval for the true mean length of rods produced by this​ process is [14.45 , 15.15].

Explanation:

We are given that a metal fabrication​ process, metal rods are produced to a specified target length of 15 feet. Suppose that the lengths are normally distributed.

A quality control specialist collects a random sample of 16 rods and finds the sample mean length to be 14.8 feet and a standard deviation of 0.65 feet.

Firstly, the pivotal quantity for 95% confidence interval for the true mean length of rods is given by;

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

where,
`\bar X` = sample mean length = 14.8 feet

s = sample standard deviation = 0.65 feet

n = sample of rods = 16

`\mu` = true mean

Here for constructing 95% confidence interval we have used t statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population​ mean,
`\mu` is ;

P(-2.131 <
`t_1_5` < 2.131) = 0.95 {As the critical value of t at 15 degree of

freedom are -2.131 & 2.131 with P = 2.5%}

P(-2.131 <
`(\bar X - \mu)/((s)/(√(n) ) )` < 2.131) = 0.95

P(
`-2.131 * {(s)/(√(n) ) }` <
`{\bar X - \mu}` <
`2.131 * {(s)/(√(n) ) }` ) = 0.95

P(
`\bar X-2.131 * {(s)/(√(n) ) }` <
`\mu` <
`\bar X+2.131 * {(s)/(√(n) ) }` ) = 0.95

95% confidence interval for
`\mu` = [
`\bar X-2.131 * {(s)/(√(n) ) }` ,
`\bar X+2.131 * {(s)/(√(n) ) }` ]

= [
`14.8-2.131 * {(0.65)/(√(16) ) }` ,
`14.8+2.131 * {(0.65)/(√(16) ) }` ]

= [14.45 , 15.15]

Hence, 95% confidence interval for the true mean length of rods produced by this​ process is [14.45 , 15.15].

Answer 2

95% Confidence interval: (14.4537 ,15.1463)

Explanation:

We are given the following in the question:

Population mean, μ = 15 feet

Sample mean,
`\bar{x}` = 14.8 feet

Sample size, n = 16

Alpha, α = 0.05

Sample standard deviation, σ = 0.65 feet

Degree of freedom = n - 1 = 15

95% Confidence interval:

`\bar{x} \pm t_(critical)\displaystyle(s)/(√(n))`

Putting the values, we get,

`t_(critical)\text{ at degree of freedom 15 and}~\alpha_(0.05) = \pm 2.1314`

`14.8 \pm 2.1314((0.65)/(√(16)) ) \\\\= 14.8 \pm 0.3463 = (14.4537 ,15.1463)`

is the required confidence interval for the true mean length of rods.