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# Testing a Claim About a Population Standard Deviation or Variance

## Assume that the population is normally distributed and that the sam...

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### Testing a Claim About a Population Standard Deviation or Variance

Assume that the population is normally distributed and that the sample has been randomly selected.

Listed below are measured amounts of lead (in micrograms per cubic meter, or ?g/m 3
?g/m3
) in the air. The Environmental Protection Agency has established an air quality standard for lead of 1.5 ?g/m 3
?g/m3
. The measurements shown below were recorded at Building 5 of the World Trade Center site on different days immediately following the destruction caused by the terrorist attacks of September 11, 2001. Test the claim that these amounts are from a population with a standard deviation greater than 0.4 ?g/m 3
?g/m3
. Use a significance level of 0.05.

0.90 1.10 0.42 0.73 0.48 1.10

b. Critical value(s): ? 2
?2

d. Initial Answerrejectfail to reject the null hypothesis.

e. Final We Answerhavedo not have enough evidence to support the claim.

Step 1 of 2

By assuming that the population is normally distributed and sample has been selected at random.
Here the measurements recorded at Building 5 of the World Trade Center site on different days immediately following the destruction caused by the terrorist attacks of September 11, 2001 is shown below.

0.90, 1.10, 0.42, 0.73, 0.48, 1.10

Now, we need to test the claim that these amounts are from a population with a standard deviation greater than 0.4 gm3.

The Hypothesis can be expressed as
H0 : = 0.4 against
H1 : 0.4

Use =0.05level of significance

The sample variance,

s = i(xi -x)2n - 1
Inserting the values 0.90, 1.10, 0.42, 0.73, 0.48, 1.10 in the scientific calculator, we get
Standard deviation s = 0.2969. Hence s2=0.0881
Given, n = 6
Degrees of freedom = n - 1
= 5

Step 2 of 2

(a)
The Test Statistics is given by,

2=(n-1)s202
Now, we need to find the calculated test statistic value
2=(n-1)s202
= 0.44070.4
= 1.1018

(b) We have given rejection region =0.05 in the upper tail of the 2 distribution with degree of freedom.
From the2table, the value of 1 - , n - 1 2=11.07
Here 2=1.1018 < 1 - , n - 1 2=11.07

(c) Since 1.1018 < 11.07, the P-value is greater than 0.05. Hence we fail to reject the null hypothesis.

(d) Therefore, we can see that the calculated values is less than the table value, hence we accept the null hypothesis at 5% level of significance.

(e) Hence there is no sufficient evidence to claim that these amounts are from a population with a standard deviation greater than 0.4 gm3.