Answer to Question #333673 in Statistics and Probability for Christine

Question

A confidence interval estimate is desired for the gain in a circuit on a semiconductor device. Assume the gain is normally distributed with standard deviation σ = 20.

a) Find a 95% confidence interval for μ when n = 10 and x̄ = 1000.

b) Find a 95% confidence interval for μ when n = 25 and x̄ = 1000.

c) Find a 99% confidence interval for μ when n = 10 and x̄ = 1000.

d) Find a 99% confidence interval for μ when n = 25 and x̄ = 1000.

Answer 1

The corresponding confidence interval is computed as shown below:

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

The critical value for $\alpha = 0.05$ is $z_c = z_{1-\alpha/2} = 1.96.$

The critical value for $\alpha = 0.01$ is $z_c = z_{1-\alpha/2} =2.5758.$

a)

$$CI=(1000-1.96\times\dfrac{20}{\sqrt{10}}, 1000+1.96\times\dfrac{20}{\sqrt{10}})$$

$$=(987.604,1012.396)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $987.604 < \mu < 1012.396,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(987.604,1012.396).$

b)

$$CI=(1000-1.96\times\dfrac{20}{\sqrt{25}}, 1000+1.96\times\dfrac{20}{\sqrt{25}})$$

$$=(992.16,1007.84)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $992.16 < \mu < 1007.84,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(992.16, 1007.84).$

c)

$$CI=(1000-2.5758\times\dfrac{20}{\sqrt{10}}, 1000+2.5758\times\dfrac{20}{\sqrt{10}})$$

$$=(983.709,1016.291)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $983.709 < \mu < 1016.291,$ which indicates that we are 99% confident that the true population mean $\mu$ is contained by the interval $(983.709,1016.291).$

d)

$$CI=(1000-2.5758\times\dfrac{20}{\sqrt{25}}, 1000+2.5758\times\dfrac{20}{\sqrt{25}})$$

$$=(989.697,1010.303)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $989.697 < \mu < 1010.303,$ which indicates that we are 99% confident that the true population mean $\mu$ is contained by the interval $(989.697,1010.303).$