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Q.8.8

Expert-verified

Introductory Statistics

Found in: Page 459

Book edition OER 2018

Author(s) Barbara Illowsky, Susan Dean

Pages 902 pages

ISBN 9781938168208

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Short Answer

You do a study of hypnotherapy to determine how effective it is in increasing the number of hours of sleep subjects get each night. You measure hours of sleep for $12$ subjects with the following results. Construct a $95$ % confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data. $8.2$ ; $9.1$ ; $7.7$ ; $8.6$ ; $6.9$ ; $11.2$ ; $10.1$ ; $9.9$ ; $8.9$ ; $9.2$ ; $7.5$ ; $10.5$

$A 95 % Cl on the population mean is 8.16 \leq μ \leq 9.80 .$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Information

Construct a $95$ % confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data. $8.2$ ; $9.1$ ; $7.7$ ; $8.6$ ; $6.9$ ; $11.2$ ; $10.1$ ; $9.9$ ; $8.9$ ; $9.2$ ; $7.5$ ;

Step 2: Explanation

If $\bar{x}$ and s are the mean and standard deviation of a random sample from a normal distribution with unknown variance $σ^{2}$ ,100(1- $α$ ) % confidence interval on is given by

$\bar{x} - t_{\frac{α}{2}, n - 1} \frac{s}{\sqrt{n}} \leq μ \leq \bar{x} + t_{\frac{α}{2}, n - 1} \frac{s}{\sqrt{n}}$ (1)

$where t_{\frac{α}{2}, n - 1} is the upper 100 \frac{α}{2} percentage point of the t distribution with n - 1 degrees of freedom.$

The simple mean is

$\bar{x} = \frac{1}{n} \sum_{i = 1}^{n} x_{i} = \frac{8.2 + 9.1 + \dots + 10.5}{12} = 8.98$

and standard deviation is

$s = {(\frac{\sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2}}{n - 1})}^{\frac{1}{2}} = {(\frac{\sum_{i = 1}^{12} {(x_{i} - 8.98)}^{2}}{11})}^{\frac{1}{2}} = 1.29$

Step 3: Explanation

$Now, we need to find a 95 % Cl on the population mean, then$

$\frac{α}{2} = \frac{1 - 0.95}{2} = 0.025 \Rightarrow t_{\frac{α}{2}, n - 1} = t_{0.025, 11} = 2.201 .$ (2)

We used a probability table for the Student's t-distribution to find the value of t. The table gives t-scores that correspond to degrees of freedom (row) and the confidence level (column). The t-score is found where the row and column intersect in the table.

From Equation (1) and (2) we get

$8.98 - 2.2 \frac{1.29}{\sqrt{12}} \leq μ \leq 8.98 + 2.2 \frac{1.29}{\sqrt{12}}$

Therefore,a $95$ % cl on the population mean is

$8.16 \leq μ \leq 9.80$

Most popular questions for Math Textbooks

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The mean age for all Foothill College students for a recent Fall term was $33.2$ . The population standard deviation has been pretty consistent at $15$ . Suppose that twenty-five Winter students were randomly selected. The mean age for the sample was $30.4$ . We are interested in the true mean age for Winter Foothill College students. Let $X =$ the age of a Winter Foothill College student.

As a result of your answer to Exercise $8.26$ , state the exact distribution to use when calculating the confidence interval.

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