Question

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population of recent Miss America winners. Use 0.01 significance level to test the claim that the BMI for recent Miss America winners are from a population with standard deviation of 1.34.

A. Identify the null hypothesis and the alternative hypothesis.
B. Find the critical value or values.
C. Find the test statistic.
D. State the conclusion that addresses the original claim.

Answer 1

a) H0:
`\sigma = 1.34`

H1:
`\sigma \\eq 1.34`

b)
`df = n-1= 10-1=9`

And the critical values with
`\alpha/2=0.005` on each tail are:

`\chi_(\alpha/2)= 1.735, \chi_(1-\alpha/2)= 23.589`

c)
`t=(10-1) [(1.186)/(1.34)]^2 =7.05`

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

`\sigma_o =1.34` the value that we want to test

`p_v` represent the p value for the test

t represent the statistic (chi square test)

`\alpha=0.01` significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0:
`\sigma = 1.34`

H1:
`\sigma \\eq 1.34`

The statistic is given by:

`t=(n-1) [(s)/(\sigma_o)]^2`

Part b

The degrees of freedom are given by:

`df = n-1= 10-1=9`

And the critical values with
`\alpha/2=0.005` on each tail are:

`\chi_(\alpha/2)= 1.735, \chi_(1-\alpha/2)= 23.589`

Part c

Replacing the info we got:

`t=(10-1) [(1.186)/(1.34)]^2 =7.05`

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34