Question

Standard deviating of the breaking strength of 10 cables tested by company was 815kg.Find 95%,and 99%confidence limit for the deviation of all cables produced by the company.

Answer 1

Standard deviating of the breaking strength of 10 cables tested by company was 815kg with standard deviation 45 kg.Find 95%,and 99%confidence limit for the deviation of all cables produced by the company.

We need to construct the 95% confidence interval for the population mean $\mu$

Since the population's $\sigma$ is not known the formula uses the T-distribution with $n-1$ degrees of freedom:

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

Given that

$$\bar{X}=815, s=45, df=n-1=10-1=9$$

For $\alpha=0.05$

$$t_c=z_ {1-\alpha/2;n-1}=2.262$$

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

$$=(815-2.262{45 \over \sqrt{10}},815+2.262\times{45 \over \sqrt{10}})=$$

$$=(782.809,847.191)$$

We need to construct the 99% confidence interval for the population mean $\mu$

For $\alpha=0.01$

$$t_c=z_ {1-\alpha/2;n-1}=3.250$$

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

$$=(815-3.250{45 \over \sqrt{10}},815+3.250\times{45 \over \sqrt{10}})=$$

$$=(768.754,861.246)$$