Confidence Interval for the Difference of Two Means

You do a survey of \(40\) small town coffee shops and \(49\) big city coffee shops, and find that the mean price of a large cup of coffee is \(\$3.75\) and in the big cities it is \(\$ 4.50\). You also know that the population standard deviation in small towns is \(1.20\), and in big cities the population standard deviation of \(0.98\).

Construct a \(99\%\) confidence interval for the difference of their two means, and draw conclusions from it.

Solution:

It helps to lay out the information you have. Call the small city Population \(1\) and the large city Population \(2\). Then you know that

\[ \begin{array}{lll} & n_1 = 40 & \bar{x}_1 = 3.75 & \sigma_1 = 1.20 \\ & n_2 = 49 & \bar{x}_2 = 4.50 & \sigma_2 = 0.98 . \end{array}\]

You know that the \(z\) critical value for a \(99\%\) confidence interval is \(2.58\). Then calculating the confidence interval for the difference in the means,

\[\begin{align} & \bar{x}_1 - \bar{x}_2 \pm (z \text{ critical value})\sqrt{\frac{\sigma_1^2}{n_1} +\frac{\sigma_2^2}{n_2} } \\ & \qquad = 3.75-4.50 \pm 2.58 \sqrt{\frac{(1.20)^2}{40} +\frac{(0.98)^2}{49} } \\ & \qquad = -0.75 \pm 2.58\sqrt{0.036 + 0.0196} \\ & \qquad \approx -0.75 \pm 0.61 \\ & \qquad = (-1.36, -0.14) .\end{align}\]

Now what can you conclude from this? First, you can conclude that the method used to construct this interval estimate is successful in capturing the actual difference in the population means about \(99\%\) of the time.

More importantly, you can conclude with \(99\%\) confidence that the actual difference in the mean price of a large cup of coffee is between \(-\$1.36\) and \(-\$0.14\). Because both endpoints of the confidence interval are negative, you can estimate that the mean price of a large cup of coffee is between \(\$0.14\) and \(\$1.36\) lower in a small town than it is in a big city.

You do a survey of \(40\) small town coffee shops and \(49\) big city coffee shops, and find that the mean price of a large cup of coffee is \(\$3.75\) and in the big cities it is \(\$ 4.50\). You also know that the sample standard deviation in small towns is \(1.00\), and in big cities the sample standard deviation of \(0.70\).

Construct a \(99\%\) confidence interval for the difference of their two means, and draw conclusions from it.

Solution:

First finding \(V_1\) and \(V_2\),

\[ \begin{align} V_1 &= \frac{s_1^2}{n_1} \\ &= \frac{1^2}{40} \\ &= 0.025 \end{align} \]

and

\[ \begin{align} V_2 &= \frac{s_2^2}{n_2} \\ &= \frac{0.70^2}{49} \\ &= 0.01, \end{align} \]

so

\[\begin{align} df &= \frac{(V_1 + V_2)^2}{\dfrac{V_1^2}{n_1-1} + \dfrac{V_2^2}{n_2-1} } \\ &= \frac{(0.025 + 0.01 )^2}{\dfrac{0.025^2}{40-1} + \dfrac{0.01^2}{49-1} } \\ &=\frac{0.001225}{\dfrac{0.000625}{39} + \dfrac{0.0001}{48} } \\ &\approx 67.6 . \end{align}\]

Most \(t\)-tables will not have \(df = 68\) in them, however a calculator will give you the appropriate \(t\)-critical value of \(2.65\).

Then calculating the confidence interval for the difference in the two population means,

\[\begin{align} \bar{x}_1 - \bar{x}_2 & \pm (t \text{ critical value})\sqrt{\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2} } \\ &\quad = 3.75-4.50 \pm (2.65)\sqrt{\frac{1^2}{40} +\frac{0.75^2}{49} } \\ &\quad \approx -0.75 \pm 0.51 \\ &\quad = (-1.26, -0.24). \end{align}\]

So you can conclude with \(99\%\) confidence that the actual difference in the mean price of a large cup of coffee is between \(-\$1.26\) and \(-\$0.24\). Because both endpoints of the confidence interval are negative, you can estimate that the mean price of a large cup of coffee is between \(\$0.24\) and \(\$1.26\) lower in a small town than it is in a big city.

Suppose you have a new medical treatment, and you want to look at the mean days to recovery of people who get the treatment versus people who don't get the treatment. People were randomly assigned to either the treatment group or a placebo group. Define

• Population \(1\) - people who get the treatment; and
• Population \(2\) - people who get a placebo.

Suppose the \(90\%\) confidence interval for the difference in the two means, \(\bar{x}_1 - \bar{x}_2 \) is \( (14.7, 23.1)\). What conclusion can you draw about whether or not the treatment versus the placebo?

Solution:

Here both of the endpoints of the confidence interval are positive. This means you think \(\mu_1 - \mu_2 > 0\), or in other words that the mean time to recovery for people who got the medical treatment is larger than the mean time to recovery for people who got the placebo, and in fact the recovery time for people who get the medical treatment is longer by at least \(14\) days. Unfortunately this would imply that the new medical treatment does not help people recover faster.

Let's use exactly the same setup as in the previous example. So

• Population \(1\) - people who get the treatment; and
• Population \(2\) - people who get a placebo.

Suppose the \(90\%\) confidence interval for the difference in the two means, \(\bar{x}_1 - \bar{x}_2 \) is \( (-3.4, 4.3)\). What conclusion can you draw about whether or not the treatment versus the placebo?

Solution:

Here zero is included in the confidence interval. That implies that it is plausible that \(\mu_1\) and \(\mu_2\) are the same. In other words, it is plausible that the new medical treatment was no more or less effective than the placebo. So you can say that while the new medical treatment probably didn't help, it also probably wasn't any worse than the placebo.

It is common for students in a class to be given a pre-test, then learn the material, then do an actual test. This is to (hopefully) measure how much students are learning in the class. Is this actually a case where you would construct a confidence interval for the difference between the two population means?

Solution:

Remember that one of the conditions to construct a confidence interval for the difference of two means is that your samples are independent. In this example, a student who takes the pre-test automatically is put in the group to take the actual test. These samples are definitely not independent!

So while it looks like a difference of two means question, in fact you will need to look at the people in the class and the difference in their test scores and do a confidence interval for a population mean.

Just because there is the word "difference" there doesn't imply you have to do a confidence interval for the difference between two means. These are considered matched paired samples, and a standard confidence interval for a population mean is the way to tackle this problem.

Suppose that you want to know if the color of the coffee mug impacts how people think about the flavor. You get \(24\) people and assign them randomly to one of two treatment groups; either a white coffee mug or an orange coffee mug.

Is coffee in an orange mug somehow better?

Both groups were given exactly the same coffee and asked to rate the flavor on a scale of \(0\) to \(100\). The results are in the table below.

Can you conclude that the color of the mug makes a difference in the mean quality rating of the coffee?

Solution:

First let's check to be sure that all of the conditions to construct a confidence interval for the difference of two means are satisfied. The samples are certainly independent and randomly selected, however the sample size is less than \(30\). That means you will need to assume that the distributions of the two quality ratings are approximately normal. It isn't unreasonable to assume that, but it will need to be mentioned when you make a conclusion.

Next you will need to calculate the degrees of freedom. Here

\[ \begin{align} V_1 &= \frac{s_1^2}{n_1} \\ &= \frac{(20.17)^2}{12} \\ & \approx 33.9, \end{align} \]

and

\[ \begin{align} V_2 &= \frac{s_2^2}{n_2} \\ &= \frac{(16.69)^2}{12} \\ &\approx 23.2, \end{align}\]

so

\[\begin{align} df &= \frac{(33.9 + 23.2)^2}{\dfrac{(33.9)^2}{12-1} + \dfrac{(23.2)^2}{12-1} } \\ &= \frac{3260.41}{\dfrac{1149.21}{11} + \dfrac{538.23}{11} } \\ &\approx 21.25 . \end{align}\]

A confidence level wasn't given, but a \(95\%\) level is common to use. So the \(t\)-critical value would be \(2.08\).

Then constructing the confidence interval,

\[ \begin{align} \bar{x}_1 - \bar{x}_2 &\pm (t \text{ critical value})\sqrt{\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2} } \\ &\quad = 50.35 - 61.49 \pm 2.08\sqrt{\frac{(20.17)^2}{12} +\frac{(16.69)^2}{12} } \\ & \quad \approx -11.14 \pm 14.72 \\ &\quad = (-26.85, 4.67).\end{align}\]

So assuming that the distributions of the two quality ratings are approximately normal, you can conclude with \(95\%\) confidence that the actual difference in the mean rating is between \(-26.85\) and \(4.67\). Because zero is in the confidence interval it is plausible to conclude that there is no difference in the mean flavor scale rating between the white mug and the orange mug.