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Wize University Statistics Textbook > Inference about a Population Variance

Hypothesis Test for Population Variance

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Hypothesis Test for Population Variance

We can conduct hypothesis tests for the population variance (standard deviation).

Examples:
  • Did the new oven reduce the variability in the time it takes to bake cookies?
  • Did the time in the elevator to get from G to PH become less variable since the repairs?
  • Did switching to web-based instruction increase how varied the course evaluation scores are for Professor Philipp's history class?

Wize Tip
Key words: variability, variance, vary, spread, inconsistency, fluctuation, disparity

Hypotheses

Null hypothesis:

Ho:σ2=σ02H_o:\sigma^2 =\sigma^2_0

Alternative hypothesis (depending on context):

Ha:σ2>σ02H_a:\sigma^2 >\sigma^2_0 ("more than"; one-sided)
Ha:σ2<σ02H_a:\sigma^2 <\sigma^2_0 ("less than"; one-sided)
Ha:σ2≠σ02H_a:\sigma^2 \neq\sigma^2_0 ( "not equal to", "differ", "changed"; two-sided)

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Test statistic:

χ2=(n−1)s2σo2\displaystyle\boxed{\chi^2=\frac{\left(n-1\right)s^2}{\sigma_o^2}}

where:

n=n= sample size

s2=s^2= sample variance

σo2=\sigma_o^2= population variance under the null hypothesis HoH_o (the "claim", status quo)


Degrees of freedom:

df=v=n−1df=v=n-1

Test Statistic vs. Critical Value

Alternative HypothesisCritical RegionHa:σ2>σ02χ2>χα2Ha:σ2<σ02χ2<χ1−α2Ha:σ2≠σ02χ2>χα22χ2<χ1−α22\begin{array}{|c|c|} \hline Alternative \space Hypothesis&Critical \space Region\\ \hline H_a:\sigma^2 >\sigma^2_0&\chi^2>\chi^2_{\alpha}\\ \hline H_a:\sigma^2 <\sigma^2_0&\chi^2<\chi^2_{1-\alpha}\\ \hline H_a:\sigma^2 \neq\sigma^2_0&\chi^2>\chi^2_\frac{\alpha}{2} \\&\chi^2<\chi^2_{1-\frac{\alpha}{2}} \\ \hline \end{array}


  • If χ2\chi^2 is inside the critical region →\rightarrow Reject HoH_o
  • If χ2\chi^2 is outside the critical region →\rightarrow Fail to reject HoH_o

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P-value vs. Significance Level

  • If p-value <α→< \alpha \rightarrow Reject HoH_o
  • If p-value >α→> \alpha \rightarrow Fail to reject HoH_o


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Example:

If v=10v=10 and χ2=19.58,\chi^2=19.58, find the p-value.

Without software, we can only find the range of the p-value using the Chi-square table:

0.025 < p-value < 0.05

Given this range of p-value, this means we can reject HoH_o if the significance level is α=0.10\alpha=0.10 or α=0.05\alpha=0.05 but not if α=0.01\alpha=0.01.


Exact p-value (using software):

The p-value is =1−P(X102≤19.58)=1−0.9665=0.0335=1-P\left(X_{10}^2\le19.58\right)=1-0.9665=0.0335
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Example: Hypothesis Test for Population Variance


Helen is the Customer Service Assistant Manager who supervises the customer service reps (CSR). Last year, the variance for the time it takes a CSR to resolve customer inquires on the phone is 9 minutes. Helen is frustrated because she believes the variation has increased since the company hired a bunch of new trainees. A random sample of 41 phone inquiries revealed a mean of 7.5 minutes with a standard deviation of 3.6 minutes.

Helen assumes the data is normally distributed.

(a) State the hypotheses.

Ho:σ2=9H_o:\sigma^2=9 “Variation has not increased.”
Ha:σ2>9H_a:\sigma^2>9 “Variation has increased.”


(b) Is the critical region on the left tail or the right tail of the Chi-square distribution?

Right tail because we are testing for "increased"


df=v=41−1=40df=v=41-1=40

s=3.6s=3.6
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(c) Find the critical value at the 1% significance level.

Alternative HypothesisCritical RegionHa:σ2>σ02χ2>χα2Ha:σ2<σ02χ2<χ1−α2Ha:σ2≠σ02χ2>χα22χ2<χ1−α22\begin{array}{|c|c|} \hline Alternative \space Hypothesis&Critical \space Region\\ \hline H_a:\sigma^2 >\sigma^2_0&\chi^2>\chi^2_{\alpha}\\ \hline H_a:\sigma^2 <\sigma^2_0&\chi^2<\chi^2_{1-\alpha}\\ \hline H_a:\sigma^2 \neq\sigma^2_0&\chi^2>\chi^2_\frac{\alpha}{2} \\&\chi^2<\chi^2_{1-\frac{\alpha}{2}} \\ \hline \end{array}
Critical value based on α=0.01, v=40\alpha=0.01,\ v=40

Since we are testing for Ha:σ2>σ02H_a:\sigma^2 >\sigma^2_0, the critical value is χ0.01,402=63.691\chi^2_{0.01,40}=63.691


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(d) Solve for the test statistic.

s=3.6s=3.6
X2=(n−1)s2σo2=(41−1)(3.6)2(3)2=57.6X^2=\frac{(n−1)s^2}{\sigma_o^2}=\frac{\left(41-1\right)\left(3.6\right)^2}{\left(3\right)^2}=57.6

(e) At the 1% significance level, is there evidence that the variance for the time it takes a CSR to resolve customer inquires?


The test statistic χ2=57.6<\chi^2=57.6< the critical valueχ0.01,402=63.691\chi^2_{0.01,40}=63.691.
→ \rightarrow Fail to reject HoH_o

There is no evidence that the variation in the time spent to solve phone inquiries has increased.



(f) At the 5% significance level, is there evidence that the variance for the time it takes a CSR to resolve customer inquires?


The test statistic χ2=57.6>\chi^2=57.6> the critical valueχ0.05,402=55.758\chi^2_{0.05,40}=55.758.
→ \rightarrow Reject HoH_o

There is evidence that the variation in the time spent to solve phone inquiries has increased.



Practice: Hypothesis Test for Population Variance

Homer works in a power plant. Since his job is so important, Mr. Burns can’t afford to have Homer be late for work. He asks Smithers to monitor Homer’s punctuality. A random sample of 30 shifts reveal that Homer is late for an average of 0 minutes* with a standard deviation of 12.6 minutes. Assume a normal distribution.

(*If Homer is -5 minutes late, that actually means he's 5 minutes early.)

Can Smithers conclude that the variance in Homer’s punctuality is greater than 100 minutes?

(a) What is the correct set of hypotheses?
Ho:σ2=64H_o:\sigma^2 =64
Ha:σ2≠64\\H_a:\sigma^2 \neq64


n=25n=25
α=0.05\alpha=0.05
s=5.5s=5.5

Click on 'Hint' to see formula and Chi-square Table.
(a) What are the critical values?