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:
Given that
"\\bar{X}=815, s=45, df=n-1=10-1=9"For "\\alpha=0.05"
"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"
"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)"
Comments
Leave a comment