Question

A quality control engineer wishes to estimate the mean output voltage of a DC power supply for different loads. The engineer measures output voltage when the supply is connected to 11 different loads and computes a mean 98.4 V and standard deviation 7 V. Assume that the output voltages follow a normal distribution.

(a) Construct and interpret the 99.9% confidence interval.

(b) Construct and interpret the 90% prediction interval.

Answer 1

The provided sample mean is 98.4 and the sample standard deviation is s = 7 . The size of the sample is n = 11 and the required confidence level is 99.9%.

The number of degrees of freedom are df = 11 - 1 = 10 , and the significance level is α=0.000999

Based on the provided information, the critical t-value for α=0.000999 and df = 10 degrees of freedom is t_c = 4.587

The 99.9% confidence for the population mean μ is computed using the following expression

Therefore, based on the information provided, the 99.9 % confidence for the population mean μ is

We can be 99.9% sure that true value of mean lie in the interval ( 88.719, 108.081 )

The provided sample mean is 98.4 and the sample standard deviation is s = 7 . The size of the sample is n = 11 and the required confidence level is 90%.

The number of degrees of freedom are df = 11 - 1 = 10 , and the significance level is α=0.1.

Based on the provided information, the critical t-value for α=0.1 and df = 10 degrees of freedom is t_c = 1.812

The 90% confidence for the population mean μ is computed using the following expression

Therefore, based on the information provided, the 90 % confidence for the population mean \muμ is

CI = (94.575, 102.225)

We can be 90% sure that true value of mean lie in the interval (94.575,102.225 )