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A lab technician is tested for her consistency by taking multiple measurements of cholesterol levels from the same blood sample. The target accuracy is a standard deviation of 1.2 or less.

If 20 measurements are taken and their standard deviation is 1.8, is there enough evidence to reject the claim that her standard deviation is within the limit at alfa= .01?

A) No, comparing the test value 45.00 and the critical value 37.566.

B) Yes, comparing the test value 42.75 and the critical value 36.191.

C) Yes, comparing the test value 45.00 and the critical value 37.566.

D) No, comparing the test value 42.75 and the critical value 36.191.

Answer :

Answer:

Correct option: B) Yes, comparing the test value 42.75 and the critical value 36.191.

Step-by-step explanation:

A hypothesis test is conducted to determine whether the consistency of a lab technician has a standard deviation greater than 1.2.

Then the variance will be, [tex]\sigma^{2}=(1.2)^{2}=1.44[/tex]

The hypothesis to determine whether the population variance is greater than 1.44 is:

H₀: The population variance is not less than 25.0 minutes, i.e. σ² ≤ 1.44.

Hₐ: The population variance is less than 25.0 minutes, i.e. σ² > 1.44.

Given:

n = 20

s = 1.8

The test statistics is:

[tex]\chi ^{2}_{cal.}=\frac{(n-1)s^{2}}{\sigma^{2}}=\frac{(20-1)\times(1.8)^{2}}{(1.2)^{2}}=42.75[/tex]

Decision rule:

If the calculated value of the test statistic is greater than the critical value, [tex]\chi^{2}_{(\alpha), (n-1)}[/tex] then the null hypothesis will be rejected.

Compute the critical value as follows:

[tex]\chi^{2}_{(\alpha), (n-1)}=\chi^{2}_{(0.01),(20-1)}=\chi^{2}_{0.01, 19}=36.191[/tex]

*Use a chi-square table.

The test statistic[tex]\chi^{2}_{cal.}=42.75>\chi^{2}_{\alpha ,(n-1)}=36.191[/tex].

Thus, the null hypothesis is rejected at 1% level of significance.

Conclusion:

The consistency of a lab technician has a standard deviation greater than 1.2.

The correct option is (B).

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