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A random sample of n=100 observations produced a mean of \bar{x} =30 with a standard deviation of...

Question:

A random sample of n=100 observations produced a mean of {eq}\bar{x} =30 {/eq} with a standard deviation of s=7.

(a) Find a 90% confidence interval for {eq}\mu {/eq} Lower-bound: Upper-bound:

(b) Find a 95% confidence interval for {eq}\mu {/eq} Lower-bound: Upper-bound:

(c) Find a 99% confidence interval for {eq}\mu {/eq} Lower-bound: Upper-bound:

Confidence Interval

The confidence interval is used to estimate true value of population parameter. The formula for confidence interval is given as,

{eq}\bar x \pm {t_{\alpha /2}} \cdot \dfrac{s}{{\sqrt n }}{/eq}

Here,

{eq}\bar x{/eq} is sample mean.

{eq}{t_{\alpha /2}}{/eq} is the critical value at level of significance {eq}\alpha{/eq} and degree of freedom {eq}n ? 1{/eq}.

Answer and Explanation:


Given Information:

The total number of observation is, {eq}n = 100{/eq},

The sample mean is, {eq}\bar X = 30{/eq},

The standard deviation is, {eq}s = 7{/eq},


(a)

The sample size is small, and the population variance is not known, using t-distribution,

The lower limit of 90% confidence interval is given as,

{eq}\begin{align*} \bar x - {t_{0.05,99}} \cdot \dfrac{s}{{\sqrt n }} &= 30 - \left( {1.29} \right) \cdot \dfrac{7}{{\sqrt {99} }}\\ & = 29.09 \end{align*}{/eq}

The upper limit of 90% confidence interval is given as,

{eq}\begin{align*} \bar x + {t_{0.05,99}} \cdot \dfrac{s}{{\sqrt n }} &= 30 + \left( {1.29} \right) \cdot \dfrac{7}{{\sqrt {99} }}\\ & = 30.91 \end{align*}{/eq}

Therefore, the 90% confidence interval is {eq}\left( {29.09,30.91} \right){/eq}.

(b)

The lower limit of 95% confidence interval is given as,

{eq}\begin{align*} \bar x - {t_{0.025,99}} \cdot \dfrac{s}{{\sqrt n }} &= 30 - \left( {1.984} \right) \cdot \dfrac{7}{{\sqrt {99} }}\\ & = 28.604 \end{align*}{/eq}

The upper limit of 95% confidence interval is given as,

{eq}\begin{align*} \bar x + {t_{0.025,99}} \cdot \dfrac{s}{{\sqrt n }} &= 30 + \left( {1.984} \right) \cdot \dfrac{7}{{\sqrt {99} }}\\ &= 31.396 \end{align*}{/eq}

Therefore, the 95% confidence interval is {eq}\left( {28.604,31.396} \right){/eq}.

(c)

The lower limit of 99% confidence interval is given as,

{eq}\begin{align*} \bar x - {t_{0.005,99}} \cdot \dfrac{s}{{\sqrt n }} &= 30 - \left( {2.626} \right) \cdot \dfrac{7}{{\sqrt {99} }}\\ &= 28.15 \end{align*}{/eq}

The upper limit of 99% confidence interval is given as,

{eq}\begin{align*} \bar x + {t_{0.005,99}} \cdot \dfrac{s}{{\sqrt n }} &= 30 + \left( {2.626} \right) \cdot \dfrac{7}{{\sqrt {99} }}\\ & = 31.85 \end{align*}{/eq}

Therefore, the 99% confidence interval is {eq}\left( {28.15,31.85} \right){/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
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