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Wize AP Statistics Textbook > Inference for Two Population Means

Unpooled t-test

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Hypothesis Test for Two Independent Means: Unpooled t-test for Unequal Variances

Our approach to comparing the two means depends if we if can assume the population variances σ1 σ_1\ and σ2σ_2 are equal or not, even if we don't know their values.

If we cannot assume that the population variances are equal →\rightarrow Unpooled t-test (Unequal Variances t-test)
σ1 ≠ σ2σ_1\ \ne\ σ_2

t=x‾1−x‾2s12n1+s22n2\displaystyle\boxed{t=\frac{\overline{x}_1-\overline{x}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}}
Degrees of freedom:

With software:
df=(s12n1+s22n2)21n1−1(s12n1)2+1n2−1(s22n2)2\displaystyle{df=\frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1}\left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1}\left(\frac{s_2^2}{n_2}\right)^2}}
Without software (and if allowed):

df=min⁡(n1−1, n2−1)df=\min\left(n_1-1,\ n_2-1\right)

Wize Concept
If you are doing a Pooled t-test, there is an additional assumption which is that σ1 = σ2σ_1\ =\ σ_2.



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Example: Hypothesis Test for Differences in Population Means (Unpooled t-test)

A company sells educational materials that should improve reading ability in elementary school pupils. A consultant arranges for 21 grade 3 students to use the educational materials for 8 weeks. A control classroom of 23 students follows the same curriculum without the company’s educational materials.

Using the unpooled/unequal variance method, test the hypothesis that the mean reading scores for the treatment group is higher than that of the control group. The populations are independent and the significance level is 5%. The following shows the results of reading ability scores:
  • xˉ1 =51.48, xˉ2=41.52x̄_1\ =51.48,\ x̄_2=41.52
  • s1=11.01 , s2=17.15s_1=11.01\ ,\ s_2=17.15
  • n1=21 , n2=23n_1=21\ ,\ n_2=23
(a) State the hypotheses

Based on the keyword "higher", this is a one-sided test.

Ho: μ1−μ2=0H_o:\ \mu_1-\mu_2=0
Ha: μ1−μ2>0H_a:\ \mu_1-\mu_2>0


(b) Calculate the Unpooled/Unequal Variance test statistic and its degrees of freedom:
t=x‾1−x‾2s12n1+s22n2\displaystyle\boxed{t=\frac{\overline{x}_1-\overline{x}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}}

t=(51.48−41.52)11.01221+17.15223=2.31\displaystyle{t=\frac{\left(51.48-41.52\right)}{\sqrt{\frac{11.01^2}{21}+\frac{17.15_{ }^2}{23}}}=2.31}

Degrees of freedom:

Method #1
df=(s12n1+s22n2)21n1−1(s12n1)2+1n2−1(s22n2)2\displaystyle{df=\frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1}\left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1}\left(\frac{s_2^2}{n_2}\right)^2}}

df=(11.01221+17.15223)2121−1(11.01221)2+123−1(17.15223)2=37.86 ∼37df=\frac{\left(\frac{11.01^2}{21}+\frac{17.15^2}{23}\right)^2}{\frac{1}{21-1}\left(\frac{11.01^2}{21}\right)^2+\frac{1}{23-1}\left(\frac{17.15^2}{23}\right)^2}=37.86\ \sim37

Method #2

df=min⁡(n1−1, n2−1)df=\min\left(n_1-1,\ n_2-1\right)

df=min⁡(21−1,23−1)=min⁡(20,22)=20df=\min\left(21-1,23-1\right)=\min\left(20,22\right)=20


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(c) Find the range of p-values for each method. (Use the closest available dfdf provided in the table.)



t=2.31, df=37 →\rightarrow30 or 40:
  • From t-table, 0.01<0.01< p-value <0.025<0.025
t=2.31, df=20:
  • From t-table, 0.01<0.01< p-value <0.025<0.025
In both methods, 0.01<0.01< p-value <0.025<0.025
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(d) Draw your conclusion. Are reading scores for the treatment group is higher than that of the control group?

  • pvalue<0.05 =>pvalue ≤αpvalue<0.05\ =>pvalue\ \leα therefore we reject HoH_o.
  • At the 5% level of significance, we have enough evidence to reject HoH_o.
  • The mean reading score of the treatment group is significantly greater than the mean reading score of the control group.
In a random sample of 18 non-smokers (Group 1), they lived on average 69 years with a standard deviation of 4.4 years. In a random sample of 20 smokers (Group 2), they lived on average 66 years with a standard deviation of 2 years.
Assumptions:
  • Both populations are normal
  • The population variances are not assumed to be equal
At the 1% level of significance, we want to determine if there is evidence that non-smokers live longer than smokers.

(i) State the hypotheses.