Answered: A random sample of 18 health…

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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Question

A random sample of 18 health maintenance organizations (HMO) were selected. For each HMO, the copayment in dollars for a doctors office visit was recorded. The results are as follows.

53,38,48,42,28,25,38,27,29,53,29,54,25, 36,41,53,56,56

under the assumption that copayment amounts are normally distributed, find a 90% confidence interval for the mean copayment a mountain dollars. Give the lower limit and upper limit of a 90% confidence interval.

Carrier intermediate computation so at least three decimal places. Round your answer to one decimal place.

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

Given:

For each random sample of 18 health maintenance organizations (HMO), the copayment in dollars for a doctors office visit was recorded. The results are as follows.

53,38,48,42,28,25,38,27,29,53,29,54,25, 36,41,53,56,56

Sample size,

$n = 18$

We need to construct 90% confidence interval for the mean copayment.

Here the sample size $n < 30$ and the population standard deviation $σ$ is not known , so we use $t$ distribution for finding out the critical value.

The confidence interval for population mean is given by;

$C I = x \pm t_{α / 2} \times \frac{s}{\sqrt{n}} where x is the sample mean t_{α / 2} is the critical value s is the sample standard deviation n is the sample size$