Answered: sample of size 1045 with a mean of 33.5…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Assume that a sample is used to estimate a population mean μμ. Find the 99.9% confidence interval for a sample of size 1045 with a mean of 33.5 and a standard deviation of 13.5.

Assume that a sample is used to estimate a population mean $ \mu $. Find the 99.9% confidence interval for a sample of size 1045 with a mean of 33.5 and a standard deviation of 13.5. Enter your answer as a tri-linear inequality accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

\[ \text{[ ]} < \mu < \text{[ ]} \]

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Expert Solution

Step 1

As the population standard deviation is not known, we will use t-distribution.

Given:

$X ˉ = 33.5$

$s = 13.5$ .

$n = 1045$

Confidence level is= 99.9%

The number of degrees of freedom will be

$d f = 1045 - 1 = 1044$

The significance level will be,

$α = 0.00099999999999989$ .

Based on the provided information, the critical t-value for $α = 0.001$ and $d f = 1044$ degrees of freedom is 3.29987, i.e., $3 . 300 (rounded to three decimal places)$ .

The 99.9% confidence can be calculated as:

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