A random sample of size 15 is taken from a population assumed to be normal, and x = 1.2 and s =...

Question:

A random sample of size 15 is taken from a population assumed to be normal, and {eq}\bar{x} {/eq} = 1.2 and {eq}s {/eq} = 0.6. Calculate a 98 percent confidence interval for {eq}\mu {/eq}.

Confidence interval :


The confidence interval is the specified range of values in which the true population parameter will lie.

The t-test is applied when the standard deviation of the population is not known and the sample size is less than 30.

Answer and Explanation:


Given Information:

  • The sample size (n) is 15.
  • The confidence level (CL) is 0.98.
  • The sample mean{eq}\left( {\bar x} \right) {/eq}is 1.2.
  • The sample standard deviation (s) is 0.6.

The formula of the confidence interval for one-sample t-test is:

{eq}\bar x \pm \left( {{t^ * } \times \dfrac{s}{{\sqrt n }}} \right) {/eq}

The level of significance {eq}\left( \alpha \right) {/eq}is:

{eq}\begin{align*} \alpha &= 1 - CL\\ &= 1 - 0.98\\ &= 0.02 \end{align*} {/eq}

The degrees of freedom (df) is:

{eq}\begin{align*} df &= n - 1\\ &= 15 - 1\\ &= 14 \end{align*} {/eq}

At the significance level 0.02 and the degrees of freedom 14, the two tailed critical value {eq}\left( {{t^ * }} \right) {/eq}obtained from the t-table is {eq}\pm 2.624 {/eq}.

The 98% confidence interval for {eq}\mu {/eq} is:

{eq}\begin{align*} \bar x \pm \left( {{t^ * } \times \dfrac{s}{{\sqrt n }}} \right) &= 1.2 \pm \left( {2.624 \times \dfrac{{0.6}}{{\sqrt {15} }}} \right)\\ &= 1.2 \pm 0.4065\\ & = \left( {0.7935,1.6065} \right) \end{align*} {/eq}

Therefore, the 98% confidence interval for {eq}\mu {/eq} is{eq}\left( {0.7935,1.6065} \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
6.2K

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