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Wize University Statistics Textbook > Inference for Two Population Means

Confidence Interval Pooled t

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Confidence Interval for Two Independent Means: Pooled t (Equal 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 can assume that the population variances are equal →\rightarrow Pooled (Equal Variances) Confidence Interval
σ1=σ2σ_1=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 equal variances is:


(x‾1−x‾2)±(t∗)(sp)1n1+1n2\boxed{\left(\overline{x}_1-\overline{x}_2\right)\pm (t^{\ast })(s_p)\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}
where,
  • t⋆t^\star is the multiplier or critical value from the t-distribution with a given degrees of freedom.

The pooled standard deviation sp\colorFour {s_p} is:
sp=(n1−1)s12+(n2−1)s22n1+n2−2s_p=\sqrt{\frac{\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2}{n_1+n_2-2}}

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Degrees of freedom:
df=n1+n2−2df=n_1+n_2-2


Wize Concept
If you are not able to assume that σ1 = σ2σ_1\ =\ σ_2, then it is more appropriate to construct an Unpooled Confidence Interval,


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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 .

df=n1+n2−2df=n_1+n_2-2
df=7+9−2=14df=7+9-2=14

Use the t-table to find t∗t^*:


t∗=1.761 t^*=1.761

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Example: Confidence Interval for Two Means (Pooled t)

17 students that don't have part-time jobs (Group 1) report studying on average 5 hours a day with a standard deviation of 1.40 hours. 10 other students that have part-time jobs (Group 2) report studying on average 4 hours a day with a standard deviation of 2.05 hours.

Assume both populations are normal and the population variances are equal.

(a) Compute a 95% confidence interval for the mean difference in hours of studying.

xˉ1 =5, xˉ2=4x̄_1\ =5,\ x̄_2=4
s1=1.4 , s2=2.05s_1=1.4\ ,\ s_2=2.05
n1=17 , n2=10n_1=17\ ,\ n_2=10

Find the degrees of freedom and then find the critical value t∗t^*:
df=n1+n2−2=17+10−2=25df=n_1+n_2-2=17+10-2=25

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Using the t-table, we get t∗=2.060t^* = 2.060

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(x‾1−x‾2)±(t∗)(sp)1n1+1n2\boxed{\left(\overline{x}_1-\overline{x}_2\right)\pm (t^{\ast })(s_p)\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

sp=(n1−1)s12+(n2−1)s22n1+n2−2s_p=\sqrt{\frac{\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2}{n_1+n_2-2}}

sp=(17−1)(1.4)2+(10−1)(2.05)217+10−2=1.6635s_p=\sqrt{\frac{\left(17-1\right)\left(1.4\right)^2+\left(10-1\right)\left(2.05\right)^2}{17+10-2}}=1.6635

(5−4)±(2.060)(1.6635)117+110\left(5-4\right)\pm\left(2.060\right)\left(1.6635\right)\sqrt{\frac{1}{17}+\frac{1}{10}}

1±1.371\pm 1.37

[−0.37 , 2.37]\left[-0.37\ ,\ 2.37\right]

(b) Interpret your results.

We are 95% confident that students without part-time jobs study on average between 0.37 hours less and 2.37 hours more than students that have part-time jobs.

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

The confidence interval contains 0. There is no evidence that the means differ.


Two simple random samples were collected to compare the average spending of Blue Jay fans (Group 1) and Yankee fans (Group 2) at baseball games.



Assume that the two populations have equal variances. Construct a 95% confidence interval.
(i) What is the point estimate?
Extra Practice
Confidence interval and Hypothesis testing
Mr. Emin loves going to Starbucks drive-thru on his busy mornings but he can’t help feeling frustrated by the amount of time he has to wait to be served. He decides to go to Starbucks for 5 days and Tim Hortons for 7 days. He records the length of time, rounded to the nearest minute he has to wait each day to be served. Here is the data
The standard deviation for Starbucks waiting time is 3.082.
The standard deviation for Tim Hortons waiting time is 2.160.
Confidence interval pooled t-test
Zelda runs coffee shops in Burnaby. She wants to compare the sales in her Metrotown location and the sales in her Brentwood Mall location. Days are randomly sampled. Summary:
Sample SizeMean Sales ($)Standard DeviationMetrotown61$460159Brentwood Mall61$33592\begin{array}{|c|c|c|c|}\hline &\text{Sample Size} & \text{Mean Sales (\$)} & \text{Standard Deviation}\\\hline \text{Metrotown} & 61 &\$460& 159\\\hline \text{Brentwood Mall}& 61& \$335 & 92\\\hline \end{array}
Construct a 90% confidence interval for the difference in the true mean sales. Assume equal population variances.