9.3 Statistical Inference for Two Population Means with Unknown Population Standard Deviations

A company that manufacturers and services photocopiers wants to study the difference in the average repair time for the two different models of photocopiers they make.  In a sample of 60 repairs of photocopier A, the mean repair time was 84.2 minutes with a standard deviation of 19.4 minutes.  In a sample of 70 repairs of photocopier B, the mean repair time was 91.6 minutes with a standard deviation of 18.8 minutes.

1. Construct a 95% confidence interval for the difference in the mean repair time for the two photocopiers.
2. Interpret the confidence interval found in part 1.
3. Is there evidence to suggest that the mean repair times for the photocopiers is the same?  Explain.

Solution:

1. Let photocopier A be population 1 and photocopier B be population 2.  These populations are independent because there is no relationship between the repair times for the two photocopiers. From the question we have the following information:
   

   Photocopier A	Photocopier B
   [latex]n_1=60[/latex]	[latex]n_2=70[/latex]
   [latex]\overline{x}_1=84.2[/latex]	[latex]\overline{x}_2=91.6[/latex]
   [latex]s_1=19.4[/latex]	[latex]s_2=18.8[/latex]

   To find the confidence interval, we need to find the [latex]t[/latex]-score for the 95% confidence interval.  This means that we need to find the [latex]t[/latex]-score so that the area in the tails is [latex]1-0.95=0.05[/latex].

   [latex]\begin{eqnarray*} \\ df  & = &  \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1} \times \left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1} \times \left(\frac{s_2^2}{n_2}\right)^2} \\    & = &  \frac{\left(\frac{19.4^2}{60}+\frac{18.8^2}{70}\right)^2}{\frac{1}{60-1} \times \left(\frac{19.4^2}{60}\right)^2+\frac{1}{70-1} \times \left(\frac{18.8^2}{70}\right)^2} \\ & = & 123.68.... \\ & \Rightarrow & 123 \\ \\ \end{eqnarray*}[/latex]

   

   Function	t.inv.2t	Answer
   Field 1	0.05	1.9794…
   Field 2	123

   So [latex]t=1.9794...[/latex]. The 95% confidence interval is

   [latex]\begin{eqnarray*} \\ \mbox{Lower Limit} & = & \overline{x}_1-\overline{x}_2-t \times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ & = & 84.2-91.6-1.9794... \times \sqrt{\frac{19.4^2}{60}+\frac{18.8^2}{70}} \\ & = & -14.06  \\ \\ \mbox{Upper Limit} & = & \overline{x}_1-\overline{x}_2+t \times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ & = & 84.2-91.6+1.9794... \times \sqrt{\frac{19.4^2}{60}+\frac{18.8^2}{70}} \\ & = & -0.74 \\ \\ \end{eqnarray*}[/latex]

2. We are 95% confident that the difference in the mean repair time for the two photocopiers is between -14.06 minutes and -0.74 minutes.
3. Because 0 is outside the confidence interval and both limits are negative, it suggests that the difference in the means [latex]\displaystyle{\mu_1-\mu_2}[/latex] is less than 0.  That is, [latex]\displaystyle{\mu_1-\mu_2 \lt 0}[/latex] ([latex]\mu_1 \lt \mu_2[/latex]).  This suggests that the mean for population 1 (photocopier A) is less than the mean for population 2 (photocopier B).  So the mean repair time for photocopier A is less than the mean repair time for photocopier B.

A researcher wants to study the difference between the average amount of time boys and girls aged seven to eleven spend playing sports each day. In a sample of 9 girls, the average number of hours spent playing sports per day is 2 hours with a standard deviation of 0.866 hours.  In a sample of 16 boys, the average number of hours spent playing sports per day is 3.2 hours with a standard deviation of 1 hours.  Both populations have a normal distribution.  At the 5% significance level, is there a difference in the mean amount of time boys and girls aged seven to eleven play sports each day?

Solution:

Let girls be population 1 and boys be population 2.  These populations are independent because there is no relationship between the two groups.  From the questions, we have the following information:

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & \mu_1-\mu_2=0 \\ H_a: & & \mu_1-\mu_2 \neq 0  \end{eqnarray*}[/latex]

p-value: 

This is a test on a the difference in two population means where the population standard deviation are unknown.  So we use a [latex]t[/latex]-distribution to calculate the p-value.  Because the alternative hypothesis is a [latex]\neq[/latex], the p-value is the sum of areas in the tails of the distribution.

To use the t.dist.2t function, we need to calculate out the [latex]t[/latex]-score and the degrees of freedom:

[latex]\begin{eqnarray*} t & = & \frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ & = & \frac{(2-3.2)-0}{\sqrt{\frac{0.866^2}{9}+\frac{1^2}{16}}} \\ & = & -3.1423...\\ \\ df  & = &  \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1} \times \left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1} \times \left(\frac{s_2^2}{n_2}\right)^2} \\    & = &  \frac{\left(\frac{0.866^2}{9}+\frac{1^2}{16}\right)^2}{\frac{1}{9-1} \times \left(\frac{0.866^2}{9}\right)^2+\frac{1}{16-1} \times \left(\frac{1^2}{16}\right)^2} \\ & = & 18.846... \\ & \Rightarrow & 18 \end{eqnarray*}[/latex]

So the p-value[latex]=0.0056[/latex].

Conclusion:

Because p-value[latex]=0.0056 \lt 0.05=\alpha[/latex], we reject the null hypothesis in favour of the alternative hypothesis.  At the 5% significance level there is enough evidence to suggest that there is a difference in the mean amount of time boys and girls aged seven to eleven play sports each day.

A town has two colleges.  A local community group believes that students who graduate from College A have taken more math classes than the students who graduate from College B.  In a sample of 11 graduates from College A, the average is 4 math classes per graduate with a standard deviation of 1.5 math classes.  In a sample of 9 graduates from College B, the average is 3.5 math classes per graduate with a standard deviation of 1 math class.  Both populations have a normal distribution. At the 1% significance level, test the community groups claim that graduates from College A have taken more math classes than graduates from College B.

Solution:

Let College A be population 1 and College B be population 2.  These populations are independent because there is no relationship between the two groups.  From the questions, we have the following information:

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & \mu_1-\mu_2=0 \\ H_a: & & \mu_1-\mu_2 \gt 0  \end{eqnarray*}[/latex]

p-value: 

This is a test on a the difference in two population means where the population standard deviation are unknown.  So we use a [latex]t[/latex]-distribution to calculate the p-value.  Because the alternative hypothesis is a [latex]\gt[/latex], the p-value is the area in the right tail of the distribution.

To use the t.dist.rt function, we need to calculate out the [latex]t[/latex]-score and the degrees of freedom:

[latex]\begin{eqnarray*} t & = & \frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ & = & \frac{(4-3.5)-0}{\sqrt{\frac{1.5^2}{11}+\frac{1^2}{9}}} \\ & = & 0.8899...\\ \\  df  & = &  \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1} \times \left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1} \times \left(\frac{s_2^2}{n_2}\right)^2} \\    & = &  \frac{\left(\frac{1.5^2}{11}+\frac{1^2}{9}\right)^2}{\frac{1}{11-1} \times \left(\frac{1.5^2}{11}\right)^2+\frac{1}{9-1} \times \left(\frac{1^2}{9}\right)^2} \\ & = & 17.397... \\ & \Rightarrow & 17 \end{eqnarray*}[/latex]

So the p-value[latex]=0.1930[/latex].

Conclusion:

Because p-value[latex]=0.1930 \gt 0.01=\alpha[/latex], we do not reject the null hypothesis.  At the 1% significance level there is not enough evidence to suggest that, on average, graduates of College A take more math classes than graduates of College B.

A professor at a large community college taught both an online section and a face-to-face section of his statistics course.  The professor wants to study the difference in the average score on the final exam, believing that the mean score for the online section would be lower than the face-to-face section.  The professor randomly selected 30 final exam scores from each section and recorded the scores in the tables below.

Online Section:

Face-to-Face Section:

At the 5% significance level, is the mean of the final exam score for the online section lower than the mean of the final exam score for the face-to-face section?

Solution:

Let the online section be population 1 and the face-to-face section be population 2.  These populations are independent because there is no relationship between the two groups.  From the questions, we have the following information:

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & \mu_1-\mu_2=0 \\ H_a: & & \mu_1-\mu_2 \lt 0  \end{eqnarray*}[/latex]

p-value: 

This is a test on a the difference in two population means where the population standard deviation are unknown.  So we use a [latex]t[/latex]-distribution to calculate the p-value.  Because the alternative hypothesis is a [latex]\lt[/latex], the p-value is the area in the left tail of the distribution.

To use the t.dist function, we need to calculate out the [latex]t[/latex]-score and the degrees of freedom:

[latex]\begin{eqnarray*} t & = & \frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ & = & \frac{(72.85-84.98)-0}{\sqrt{\frac{16.918...^2}{30}+\frac{11.714...^2}{30}}} \\ & = & -3.228...\\ \\ df  & = &  \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1} \times \left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1} \times \left(\frac{s_2^2}{n_2}\right)^2} \\    & = &  \frac{\left(\frac{16.918...^2}{30}+\frac{11.714...^2}{30}\right)^2}{\frac{1}{30-1} \times \left(\frac{16.918...^2}{30}\right)^2+\frac{1}{30-1} \times \left(\frac{11.714...^2}{30}\right)^2} \\ & = & 51.608... \\ & \Rightarrow & 51 \end{eqnarray*}[/latex]

So the p-value[latex]=0.0011[/latex].

Conclusion:

Because p-value[latex]=0.0011 \lt 0.05=\alpha[/latex], we do reject the null hypothesis in favour of the alternative hypothesis.  At the 5% significance level there is enough evidence to suggest that the mean final exam score for the online section is lower than the face-to-face section.