### Chapter 8. Testing for Differences in Mean and Proportion: Paired Samples t-test

### Confidence Interval for a Mean Difference

Confidence Interval for a Population Mean Difference

Assuming the *sampling distribution of the sample mean difference *is (approximately) normal, the general formula for computing a #C\%\,CI# for a population mean difference #\mu_D#, based on a random sample of #n# difference scores, is:

\[CI_{\mu_D}=\bigg(\bar{D} - t^*\cdot \cfrac{s_D}{\sqrt{n}},\,\,\,\, \bar{D} + t^*\cdot \cfrac{s_D}{\sqrt{n}} \bigg)\]

Where #t^*# is the *critical value *of the #t_{n-1}# distribution such that #\mathbb{P}(-t^* \leq t \leq t^*)=\frac{C}{100}#.

Calculating t* with Statistical Software

Let #C# be the *confidence level *in #\%#.

To calculate the *critical value* #t^*# in Excel, make use of the function **T.INV()**:

\[=\text{T.INV}((100+C)/200, n \text{ - } 1)\]

To calculate the *critical value* #t^*# in R, make use of the function **qt()**:

\[\text{qt}(p=(100+C)/200, df=n \text{ - } 1,lower.tail = \text{TRUE})\]

A researcher conducts an experiment in which #12# randomly selected students are invited to eat dinner at a restaurant on two different evenings. On one evening each student receives a regular-size plate and on the other, they receive a large-size plate.

On each occasion, the students are allowed to choose as much food as they want from a buffet. Once the students have made their selection, their plates are weighed.

The table below shows how much food (in grams) each student chose when they were given a regular-size plate #(X)# and when they were given a large-size plate #(Y)#:

\[\begin{array}{|l|c|c|c|c|c|c|c|c|c|c|c|c|}\hline \text{Student} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12\\ \hline X \text{: Regular} &442&335&388&327&419&449&447&300&445&320&416&358\\ \hline Y \text{: Large} &474&387&430&309&465&552&498&370&475&298&380&373\\ \hline \end{array}\]

You may assume that the amount of food eaten for either plate size is normally distributed.

Define #D=Y-X# and construct a #91\%# confidence interval for the population mean difference #\mu_D#. Round your answers to #3# decimal places.

#CI_{\mu,\,91\%}=(8.805,\,\,\, 52.029)#

There are a number of different ways we can compute the *confidence interval*. Click on one of the panels to toggle a specific solution.

Compute the difference scores:

\[\begin{array}{|l|c|c|c|c|c|c|c|c|c|c|c|c|}\hline \text{Student} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12\\ \hline X \text{: Regular} &442&335&388&327&419&449&447&300&445&320&416&358\\ \hline Y \text{: Large} &474&387&430&309&465&552&498&370&475&298&380&373\\ \hline D \text{: Difference} &32&52&42&-18&46&103&51&70&30&-22&-36&15\\ \hline \end{array}\]

Assuming the amount of food eaten for either plate size is normally distributed, we know that the *sampling distribution of the sample mean difference *is normally distributed as well.

If the *sampling distribution of the sample mean difference *is (approximately) normal, the general formula for computing a #C\%\,CI# for a population mean difference #\mu_D#, based on a random sample of size #n#, is:

\[CI_{\mu_D}=\bigg(\bar{D} - t^*\cdot \cfrac{s_D}{\sqrt{n}},\,\,\,\, \bar{D} + t^*\cdot \cfrac{s_D}{\sqrt{n}} \bigg)\]

Compute the mean of the difference scores #\bar{D}#:

\[\bar{D}=\cfrac{\sum{D}}{n} = \cfrac{32+52+42-18+46+103+51+70+30-22-36+15}{12}=30.4167\]

Compute the standard deviation of the difference scores #s_{D}#:

\[\sum{D}=32+52+42-18+46+103+51+70+30-22-36+15=365\]

\[\begin{array}{rcl}\sum{D^2}&=&32^2+52^2+42^2+(-18)^2+46^2+103^2+51^2+70^2+30^2+(-22)^2\\&&+(-36)^2+15^2\\&=&28947\end{array}\]

\[s_{D}=\sqrt{\cfrac{\sum{D^2} - \cfrac{(\sum{D})^2}{n} }{n-1}}=\sqrt{\cfrac{28947 - \cfrac{365^2}{12} }{12-1}}=40.2774\]

For a given *confidence level *#C# (in #\%#), the *critical value* #t^*# of the #t_{n-1}# distribution is the value such that #\mathbb{P}(-t^* \leq t \leq t^*)=\cfrac{C}{100}#.

To calculate this critical value #t^*# in Excel, make use of the following function:

T.INV(probability, deg_freedom)

probability: A probability corresponding to the normal distribution.deg_freedom: The mean of the distribution.

Here, we have #C=91#. Thus, to calculate #t^*# such that #\mathbb{P}(-t^* \leq t \leq t^*)=0.91#, run the following command:

\[\begin{array}{c}

=\text{T.INV}((100+C)/200, n - 1)\\

\downarrow\\

=\text{T.INV}(191/200, 12 \text{ - } 1)

\end{array}\]

This gives:

\[t^* = 1.85877\]

Calculate the lower bound #L# of the confidence interval:

\[L = \bar{D} - t^* \cdot \cfrac{s_D}{\sqrt{n}} = 30.4167 - 1.85877 \cdot \cfrac{40.2774}{\sqrt{12}}=8.805\]

Calculate the upper bound #U# of the confidence interval:

\[U = \bar{D} + t^* \cdot \cfrac{s_D}{\sqrt{n}} = 30.4167 + 1.85877 \cdot \cfrac{40.2774}{\sqrt{12}}=52.029\]

Thus, the #91\%# confidence interval for the population mean difference #\mu_D# is:

\[CI_{\mu_D,\,91\%}=(8.805,\,\,\, 52.029)\]

Compute the difference scores:

\[\begin{array}{|l|c|c|c|c|c|c|c|c|c|c|c|c|}\hline \text{Student} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12\\ \hline X \text{: Regular} &442&335&388&327&419&449&447&300&445&320&416&358\\ \hline Y \text{: Large} &474&387&430&309&465&552&498&370&475&298&380&373\\ \hline D \text{: Difference} &32&52&42&-18&46&103&51&70&30&-22&-36&15\\ \hline \end{array}\]

Assuming the amount of food eaten for either plate size is normally distributed, we know that the *sampling distribution of the sample mean difference *is normally distributed as well.

If the *sampling distribution of the sample mean difference *is (approximately) normal, the general formula for computing a #C\%\,CI# for a population mean difference #\mu_D#, based on a random sample of size #n#, is:

\[CI_{\mu_D}=\bigg(\bar{D} - t^*\cdot \cfrac{s_D}{\sqrt{n}},\,\,\,\, \bar{D} + t^*\cdot \cfrac{s_D}{\sqrt{n}} \bigg)\]

Compute the mean of the difference scores #\bar{D}#:

\[\bar{D}=\cfrac{\sum{D}}{n} = \cfrac{32+52+42-18+46+103+51+70+30-22-36+15}{12}=30.4167\]

Compute the standard deviation of the difference scores #s_{D}#:

\[\sum{D}=32+52+42-18+46+103+51+70+30-22-36+15=365\]

\[\begin{array}{rcl}\sum{D^2}&=&32^2+52^2+42^2+(-18)^2+46^2+103^2+51^2+70^2+30^2+(-22)^2\\&&+(-36)^2+15^2\\&=&28947\end{array}\]

\[s_{D}=\sqrt{\cfrac{\sum{D^2} - \cfrac{(\sum{D})^2}{n} }{n-1}}=\sqrt{\cfrac{28947 - \cfrac{365^2}{12} }{12-1}}=40.2774\]

For a given *confidence level *#C# (in #\%#), the *critical value* #t^*# of the #t_{n-1}# distribution is the value such that #\mathbb{P}(-t^* \leq t \leq t^*)=\cfrac{C}{100}#.

To calculate this critical value #t^*# in R, make use of the following function:

qt(p, df, lower.tail)

p: A probability corresponding to the normal distribution.df: An integer indicating the number of degrees of freedom.lower.tail: If TRUE (default), probabilities are #\mathbb{P}(X \leq x)#, otherwise, #\mathbb{P}(X \gt x)#.

Here, we have #C=91#. Thus, to calculate #t^*# such that #\mathbb{P}(-t^* \leq t \leq t^*)=0.91#, run the following command:

\[\begin{array}{c}

\text{qt}(p = (100+C)/200, df = n \text{ - } 1, lower.tail = \text{TRUE})\\

\downarrow\\

\text{qt}(p =191/200, df = 12 \text { - } 1, lower.tail = \text{TRUE})

\end{array}\]

This gives:

\[t^* = 1.85877\]

Calculate the lower bound #L# of the confidence interval:

\[L = \bar{D} - t^* \cdot \cfrac{s_D}{\sqrt{n}} = 30.4167 - 1.85877 \cdot \cfrac{40.2774}{\sqrt{12}}=8.805\]

Calculate the upper bound #U# of the confidence interval:

\[U = \bar{D} + t^* \cdot \cfrac{s_D}{\sqrt{n}} = 30.4167 + 1.85877 \cdot \cfrac{40.2774}{\sqrt{12}}=52.029\]

Thus, the #91\%# confidence interval for the population mean difference #\mu_D# is:

\[CI_{\mu_D,\,91\%}=(8.805,\,\,\, 52.029)\]

#\phantom{0}#

Connection to Hypothesis Testing

There exists a direct connection between a *two-sided paired samples* #t#*-test* for #\mu_D# and a #(1-\alpha)\cdot 100\%# confidence interval for #\mu_D#:

- If #0# falls
*inside*the #(1 - \alpha)\cdot 100\%\,CI#, then #H_0: \mu_D=0# should not be rejected at the #\alpha# level of significance. - If #0# falls
*outside*of the #(1 - \alpha)\cdot 100\%\,CI#, then #H_0: \mu_D=0# should be rejected at the #\alpha# level of significance.

*#94\%#*confidence interval for a population mean difference #\mu_D# is #(-1.854,\,\, 0.644)#.

Suppose you use the same sample of difference scores to test #H_0: \mu_D = 0# against #H_a: \mu_D \neq 0# at the #\alpha = 0.06# level of significance.

What would be the conclusion?

Since the #94\%# confidence interval #(-1.854,\,\,0.644)# contains the value #0#, we would not reject #H_0: \mu_D = 0# at the #\alpha = 0.06# level of significance.

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