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

Answer:

95% Confidence interval: (14.4537 ,15.1463)

Step-by-step 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:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$  

Putting the values, we get,  

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

$14.8 \pm 2.1314(\dfrac{0.65}{\sqrt{16}} ) \\\\= 14.8 \pm 0.3463 = (14.4537 ,15.1463)$  

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

Answer 2

Answer:

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

Step-by-step 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. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

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 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.131) = 0.95

P( $-2.131 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ]

                                             = [ $14.8-2.131 \times {\frac{0.65}{\sqrt{16} } }$ , $14.8+2.131 \times {\frac{0.65}{\sqrt{16} } }$ ]

                                             = [14.45 , 15.15]

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