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Wize University Statistics Textbook > Inference for Two Population Means

Confidence Interval Unpooled t

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Confidence Interval for Two Independent Means: Unpooled t (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 (Unequal Variances) Confidence Interval
σ1 ≠ σ2σ_1\ \ne\ σ_2

A 100(1−α)%100\left(1-\alpha\right)\%confidence interval for the mean difference μ1−μ2\mu_1-\mu_2 of two independent populations with unequal variances is:

x‾1−x‾2±t∗s12n1+s22n2\displaystyle\boxed{\overline{x}_1-\overline{x}_2\pm t^*\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}
where,
  • t⋆t^\star is the multiplier or critical value from the t-distribution with a given degrees of freedom.


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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 Confidence Interval, there is an additional assumption which is that σ1 = σ2σ_1\ =\ σ_2.



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Solving for t∗\colorOne {t^*} :

Example
You want construct a 90% confidence interval and your sample sizes are n1=7 n_1=7\spaceand n2=9n_2=9 .

Let's use the quicker method to solve for the degrees of freedom:

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

df=min(7−1,9−1)=min(6,8)=6df=min(7-1,9-1)=min(6,8)=6

Use the t-table to find t∗t^*:


t∗=1.943 t^*=1.943

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Example: Confidence Interval for Two Means (Unpooled t)

Treadmills Inc. wants to advertise its treadmills using a regular dude in some ads (Group 1) and a muscular male model in other ads (Group 2).
  • Over the course of 10 randomly selected ads with the regular dude, on average 150 treadmills were sold with a standard deviation of 22.
  • The company also tried 12 ads with a male model. With each of those ad an average of 120 treadmills were sold with a standard deviation of 9.
Let μ1\mu_1 represent the average number of treadmills sold with the average dude in the ads in the and μ2\mu_2 represent the average number of treadmills sold with male model in the ads.

Assume the population variances are not equal.

(a) Compute a 95% confidence interval for the mean difference in treadmills sold between the two types of ads.

x‾1−x‾2±t∗s12n1+s22n2\displaystyle\boxed{\overline{x}_1-\overline{x}_2\pm t^*\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}
xˉ1 =150, xˉ2=120x̄_1\ =150,\ x̄_2=120
s1=22 , s2=9s_1=22\ ,\ s_2=9
n1=10 , n2=12n_1=10\ ,\ n_2=12

Find the degrees of freedom and then find the critical value t∗t^*:

df=min⁡(n1−1, n2−1)=min(10−1,12−1)=min(9,11)=9df=\min\left(n_1-1,\ n_2-1\right)=min(10-1,12-1)=min(9,11)=9

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Using the t-table, we get t∗=2.262t^* = 2.262

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x‾1−x‾2±t∗s12n1+s22n2\displaystyle\boxed{\overline{x}_1-\overline{x}_2\pm t^*\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}

(150−120)±(2.262)(22)210+(9)212\displaystyle{\left(150-120\right)\pm\left(2.262\right)\sqrt{\frac{\left(22\right)^2}{10}+\frac{\left(9\right)^2}{12}}}

30±16.799530\pm 16.7995

[13.2 , 46.80]\left[13.2\ ,\ 46.80\right]

(b) Interpret your results.

We are 95% confident that on average between 13 and 47 more treadmills will be sold in the ads with the regular dude compared to the ads with the male model. (

(c) Does the confidence interval contain 0? What does that mean?

The confidence interval does not contain 0. There is evidence that the means differ.

Practice: Confidence Interval for Two Means (Unpooled t)


We want to see if the average amount of time spent driving on a bridge differs between Bridge A and Bridge B. Results:

Sample SizeMean Time on Bridge (seconds)Standard DeviationBridge A857622Bridge B617010\begin{array}{|c|c|c|c|}\hline & \text{Sample Size} & \text{Mean Time on Bridge (seconds)} & \text{Standard Deviation}\\\hline \text{Bridge A} & 85 & 76 & 22\\\hline \text{Bridge B} & 61 & 70 & 10\\\hline \end{array}
We will assume that the two populations have unequal variances.

(i) What is the point estimate?