In a hospital study, it was found that the standard deviation of the sound level

Question

In a hospital study, it was found that the standard deviation of the sound levels from 20 randomly selected areas designated "casualty doors" was 4.1 dBA and the standard deviation of 24 randomly selected areas designated as "operating theaters" was 7.5 dBA. At = 0.05, can you substantiate the claim that there is a difference in the standard deviations? Step 1: State hypotheses by filling in the symbol (=, <, >, or not equal): Ho: 1 H1: 1 Step 2: Find the critical value (from the table) (example: 2.34) Critical F value is: Step 3: Compute the test value using the formula (round to two decimal places, example 6.45, and it is always greater than 1): F test value is: Step 4: Reject the null or do not reject the null (type in either Reject the null or do not reject the null only): Step 5: Conclusion sentence (type in either is or is not only, to reflect what you found): There enough evidence to support the claim that the standard deviation of the sound levels in the two areas are different.

Explanation / Answer

Here we have to test the hypothesis that,

H0 : 1 = 2 Vs H1 : 1 2

where 1 and 2 are population standard deviations of the two populations.

Assume alpha = 0.05

The test statistics F is,

F = larger variance / smaller variance

We have given that,

n1 = larger sample size = 24

s1 = larger standard deviation = 7.5

n2 = smaller sample size = 20

s2 = smalles standard deviation = 4.1

F = 7.5^2 / 4.1^2 = 3.3462

Critical value we can find by using EXCEL.

syntax is,

=FINV(probability, deg_freedom1, deg_freedom2)

probability = alpha / 2

deg_freedom1 = n1 - 1 = 24 - 1 = 23

deg_freedom2 = n2 - 1 = 20 - 1 = 19

Critical value = 2.465

F > Critical value

Reject H0 at 5% level of significance.

Conclusion : There enough evidence to support the claim that the standard deviation of the sound levels in the two areas are different.

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