Answer :
We start by noting the sample statistics for the two independent groups:
[tex]$$
n_1=45,\quad \bar{x}_1=2.26,\quad s_1=0.8, \qquad n_2=35,\quad \bar{x}_2=2.33,\quad s_2=0.81.
$$[/tex]
Our goal is to construct a 99.9% confidence interval for the difference in means, [tex]$\mu_1 - \mu_2$[/tex]. The steps are as follows:
1. Point Estimate for [tex]$\mu_1-\mu_2$[/tex]:
The point estimate is given by the difference in the sample means:
[tex]$$
\bar{x}_1 - \bar{x}_2 = 2.26 - 2.33 = -0.07.
$$[/tex]
2. Standard Error Calculation:
Since the two samples are independent and may have different variances, the standard error (SE) for the difference is calculated as:
[tex]$$
SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}.
$$[/tex]
Substituting in the given values:
[tex]$$
SE = \sqrt{\frac{0.8^2}{45} + \frac{0.81^2}{35}} \approx 0.18.
$$[/tex]
3. Degrees of Freedom (Welch-Satterthwaite Equation):
The degrees of freedom (df) are approximated by:
[tex]$$
df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}}.
$$[/tex]
Using the data provided, this calculation gives:
[tex]$$
df \approx 72.79.
$$[/tex]
4. Finding the Critical t-Value:
For a 99.9% confidence level, the significance level is
[tex]$$
\alpha = 1 - 0.999 = 0.001.
$$[/tex]
Since the interval is two-tailed, we use [tex]$\alpha/2 = 0.0005$[/tex] in each tail. The critical t-value corresponding to this tail probability and the computed degrees of freedom is approximately:
[tex]$$
t_{\alpha/2,\,df} \approx 3.43.
$$[/tex]
5. Margin of Error:
The margin of error (MOE) is then calculated by:
[tex]$$
MOE = t_{\alpha/2,\,df} \times SE = 3.43 \times 0.18 \approx 0.62.
$$[/tex]
6. Constructing the Confidence Interval:
The 99.9% confidence interval for [tex]$\mu_1-\mu_2$[/tex] is given by:
[tex]$$
(\bar{x}_1-\bar{x}_2) \pm MOE.
$$[/tex]
Substituting our values:
[tex]$$
-0.07 \pm 0.62,
$$[/tex]
which results in the interval:
[tex]$$
(-0.69,\;0.55).
$$[/tex]
Thus, to two decimal places, the 99.9% confidence interval is:
[tex]$$
-0.07 \pm 0.62 \quad \text{or} \quad (-0.69,\;0.55).
$$[/tex]
[tex]$$
n_1=45,\quad \bar{x}_1=2.26,\quad s_1=0.8, \qquad n_2=35,\quad \bar{x}_2=2.33,\quad s_2=0.81.
$$[/tex]
Our goal is to construct a 99.9% confidence interval for the difference in means, [tex]$\mu_1 - \mu_2$[/tex]. The steps are as follows:
1. Point Estimate for [tex]$\mu_1-\mu_2$[/tex]:
The point estimate is given by the difference in the sample means:
[tex]$$
\bar{x}_1 - \bar{x}_2 = 2.26 - 2.33 = -0.07.
$$[/tex]
2. Standard Error Calculation:
Since the two samples are independent and may have different variances, the standard error (SE) for the difference is calculated as:
[tex]$$
SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}.
$$[/tex]
Substituting in the given values:
[tex]$$
SE = \sqrt{\frac{0.8^2}{45} + \frac{0.81^2}{35}} \approx 0.18.
$$[/tex]
3. Degrees of Freedom (Welch-Satterthwaite Equation):
The degrees of freedom (df) are approximated by:
[tex]$$
df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}}.
$$[/tex]
Using the data provided, this calculation gives:
[tex]$$
df \approx 72.79.
$$[/tex]
4. Finding the Critical t-Value:
For a 99.9% confidence level, the significance level is
[tex]$$
\alpha = 1 - 0.999 = 0.001.
$$[/tex]
Since the interval is two-tailed, we use [tex]$\alpha/2 = 0.0005$[/tex] in each tail. The critical t-value corresponding to this tail probability and the computed degrees of freedom is approximately:
[tex]$$
t_{\alpha/2,\,df} \approx 3.43.
$$[/tex]
5. Margin of Error:
The margin of error (MOE) is then calculated by:
[tex]$$
MOE = t_{\alpha/2,\,df} \times SE = 3.43 \times 0.18 \approx 0.62.
$$[/tex]
6. Constructing the Confidence Interval:
The 99.9% confidence interval for [tex]$\mu_1-\mu_2$[/tex] is given by:
[tex]$$
(\bar{x}_1-\bar{x}_2) \pm MOE.
$$[/tex]
Substituting our values:
[tex]$$
-0.07 \pm 0.62,
$$[/tex]
which results in the interval:
[tex]$$
(-0.69,\;0.55).
$$[/tex]
Thus, to two decimal places, the 99.9% confidence interval is:
[tex]$$
-0.07 \pm 0.62 \quad \text{or} \quad (-0.69,\;0.55).
$$[/tex]