Answer to Question #111275 in Statistics and Probability for Anonymus

Question

A machine elastic bands with breaking tension normally distributed with mean 45 N and
standard deviation 4.36 N.
a. In order to test whether there is a change in the mean breaking tension, a random
sample of 50 was tested and found to have a mean breaking tension of 43.46 N.
i. Determine a 95% confidence interval for the population mean breaking tension
based on the sample mean assuming an unchanged standard deviation.
ii. Make a conclusion for answer (a)(i).
b. If the standard deviation has changed to sigma, find the least value of sigma
for a 95 % confidence interval for the population mean to contain 45 N.

Answer 1

a. We need to construct the 95% confidence interval for the population $\mu.$ The following information is provided: $\bar{X}=43.46,\sigma=4.36,n=50,\alpha=0.05$

The critical value for $\alpha=0.05$ is $z_c=z_{1-\alpha/2}=1.96.$ The corresponding confidence interval is computed as shown below:

$CI=(\bar{X}-z_c\times{\sigma \over \sqrt{n}}, \bar{X}+z_c\times{\sigma \over \sqrt{n}})=$

$=(43.46-1.96\times{4.36 \over \sqrt{50}}, 43.46+1.96\times{4.36 \over \sqrt{50}})\approx$

$\approx(42.251,44.669)$

b. Therefore, based on the data provided, the 95% confidence interval for the population mean is $42.251<\mu<44.669,$ which indicates that we are 95% confident that the true population mean $\mu$

 is contained by the interval $(42.251,44.669).$

c.

$\bar{X}+z_c\times{\sigma \over \sqrt{n}}\geq45$

$43.46+1.96\times{\sigma \over \sqrt{50}}\geq45$

$\sigma\geq{45-43.46 \over 1.96}\times\sqrt{50}$

$\sigma\geq5.555839$

$\sigma\approx5.56$