Stats question Listed below are birth weights (in kilograms) of male babies born to mothers on aspecial vitamin supplement. Test the claim that this sample comes from a populationwith a standard deviation equal to 0.470 kg, which is the standard deviation for malebirth weights in general. Use a 0.01 significance level. Does the vitamin supplementappear to affect the variation among birth weights? In your answer, write the teststatistic, the critical values and the conclussion.3.734.373.734.333.393.684.683.523.024.092.474.134.473.223.432.54

Answered: Listed below are birth weights (in…

MATLAB: An Introduction with Applications

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ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Stats question

Listed below are birth weights (in kilograms) of male babies born to mothers on a
special vitamin supplement. Test the claim that this sample comes from a population
with a standard deviation equal to 0.470 kg, which is the standard deviation for male
birth weights in general. Use a 0.01 significance level. Does the vitamin supplement
appear to affect the variation among birth weights? In your answer, write the test
statistic, the critical values and the conclussion.
3.73
4.37
3.73
4.33
3.39
3.68
4.68
3.52
3.02
4.09
2.47
4.13
4.47
3.22
3.43
2.54

Expert Solution

Step 1

Step-by-step procedure to obtain the standard deviation using Excel is given below:

In Excel sheet, enter Birth weight in one column.
In Data, select Data Analysis and choose Descriptive statistics.
In Input Variable Range, select Birth weight.
Select Summary Statistics.
Click Labels.
Click Ok.

Output obtained using the Excel is given below:

Step 2

From the output, the sample standard deviation s is 0.6573

Step 3

Hypothesis:

The null and alternative hypothesis are below:

$\begin{array}{rcl} H_{0} : σ = 0.470 H_{1} : σ \neq 0.470 \end{array}$

The test statistic is obtained as follows:

$\begin{array}{rcl} χ^{2} & = & \frac{(n - 1) s^{2}}{σ^{2}} \\ = & \frac{(16 - 1) \times 0 . 6573^{2}}{0 . 470^{2}} \\ = & \frac{15 \times 0.4320}{0.2209} \\ = & \frac{6.4800}{0.2209} \\ = & 29.3345 \end{array}$

Thus, the test statistic is 29.3345

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