How do you solve? ref W9Q7. During the nutrition research, the amount of consumed kilocalories per day was measured for 18 people – 10 women and 8 men. Results are as follows:- Women: 1752, 1824, 1678, 1780, 1809, 1848, 1950, 1772, 1918, 1902- Men: 1911, 2181, 1846, 2251, 1894, 2492, 2019, 2170Calculate a 90% confidence interval on the mean for women and men separately. Assume distribution to be normal. Round your answers to the nearest integer (e.g. 9876).Women: [Input Box 1] ≤ μ ≤ [Input Box 2]Men: [Input Box 3] ≤ μ ≤ [Input Box 4]

Answered: How do you solve? ref W9Q7.

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

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How do you solve? ref W9Q7.

During the nutrition research, the amount of consumed kilocalories per day was measured for 18 people – 10 women and 8 men. Results are as follows:

- Women: 1752, 1824, 1678, 1780, 1809, 1848, 1950, 1772, 1918, 1902
- Men: 1911, 2181, 1846, 2251, 1894, 2492, 2019, 2170

Calculate a 90% confidence interval on the mean for women and men separately. Assume distribution to be normal. Round your answers to the nearest integer (e.g. 9876).

Women: [Input Box 1] ≤ μ ≤ [Input Box 2]

Men: [Input Box 3] ≤ μ ≤ [Input Box 4]

Expert Solution

Step 1

For Women

n_w=10

$α = 1 - 0.90 = 0.10$

Degrees of freedom is obtained as-

$d f = 10 - 1 = 9$

The mean and standard deviation is obtained as follows:

$\begin{array}{rcl} x_{w} & = & \frac{1752 + 1824 + 1678 + 1780 + . . . . + 1902}{10} \\ = & \frac{18251}{10} \\ = & 1825.1 \end{array}$

$\begin{array}{rcl} s_{w} & = & \sqrt{\frac{{(1752 - 1825.1)}^{2} + {(1824 - 1825)}^{2} + {(1678 - 1825.1)}^{2} + {(1780 - 1825.1)}^{2} + . . . . + {(1902 - 1825.1)}^{2}}{10 - 1}} \\ = & \sqrt{\frac{63048.9}{9}} \\ = & \sqrt{7005.433} \\ = & 83.6985 \end{array}$

As per t-distribution table, the critical value of t at 0.10 level of significance for 9 degrees of freedom is obtained as 1.833

The required 90% confidence interval on the mean for women is obtained as-

$\begin{array}{rcl} C I & = & x_{w} \pm t_{c r i t} \times \frac{s_{w}}{\sqrt{n}} \\ = & 1825.1 \pm (1.833) \times \frac{83.6985}{\sqrt{10}} \\ = & 1825.1 \pm 48.518 \\ = & (1776.582, 1873.618) \\ ~ (1777, 1874) \end{array}$

Women: $1777 \leq μ \leq 1874$

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