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A confidence interval estimate is desired for the gain in a circuit on a semiconductor device....

Question:

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

a) Find a 95% Cl for m when n = 10 and {eq}\bar{x} = 1200 {/eq}

b) Find a 95% CI for m when n = 25 and {eq}\bar{x} = 1200 {/eq}

c) Find a 99% CI for m when n = 10 and {eq}\bar{x} = 1200 {/eq}

d) Find a 99% Cl for m when n = 25 and {eq}\bar{x} = 1200 {/eq}

Confidence Interval:

In this question, we will use the t distribution to calculate and construct the 95% and 99% confidence interval for the population mean. A confidence interval is a point estimate of the population parameter. An excel function of t distribution TINV() is used to get the confidence coefficient.

Answer and Explanation:

a)

Given that,

  • Sample standard deviation, {eq}s = 10 {/eq}
  • Sample size, {eq}n = 10 {/eq}
  • Sample mean, {eq}\bar{x} = 1200 {/eq}


The 95% confidence interval for the population mean is defined as:

{eq}\bar{x} \pm t_{0.05/2}\frac{s}{\sqrt{n}} {/eq}


Degree of freedom, {eq}n-1 = 10 - 1 = 9 {/eq}


The excel function for the confidence coefficient:

=TINV(0.05,9)


{eq}1200 \pm 2.26\frac{20}{\sqrt{10}}\\ 1185.71 < \mu < 1214.29 {/eq}


b)

Sample size, {eq}n = 25 {/eq}

Degree of freedom, {eq}n-1 = 25 - 1 = 24 {/eq}

Excel function for the confidence coefficient:

=TINV(0.05,24)


The 95% confidence interval for the population mean is calculated as:

{eq}\bar{x} \pm t_{0.05/2}\frac{s}{\sqrt{n}}\\ 1200\pm 2.064\times \frac{20}{\sqrt{25}}\\ 1191.74 < \mu < 1208.26 {/eq}


c)

Sample size, {eq}n = 10 {/eq}

Degree of freedom, {eq}n-1 = 10 - 1 = 9 {/eq}

Excel function for the confidence coefficient:

=TINV(0.01,9)


The 99% confidence interval for the population mean is calculated as:

{eq}\bar{x} \pm t_{0.01/2}\frac{s}{\sqrt{n}}\\ 1200\pm 3.2498\times \frac{20}{10}\\ 1179.45 < \mu < 1220.55 {/eq}


d)

Sample size, {eq}n = 25 {/eq}

Degree of freedom, {eq}n-1 = 25 - 1 = 24 {/eq}

Excel function for the confidence coefficient:

=TINV(0.01,9)


The 99% confidence interval for the population mean is calculated as:

{eq}\bar{x} \pm t_{0.01/2}\frac{s}{\sqrt{n}}\\ 1200\pm 3.2498\times \frac{20}{\sqrt{25}}\\ 1187.00 < \mu < 1212.99 {/eq}


Learn more about this topic:

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Using the t Distribution to Find Confidence Intervals

from Statistics 101: Principles of Statistics

Chapter 9 / Lesson 6
6.2K

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