Question

Use the following information: \bar{x}_1=11.7, s_1=2.5, n_1=93, \bar{x}_2=10.3, s_2=2.8, n_2=66
and the formula to compute the test statistic:
t_0=(\bar{x}_1-\bar{x}_2)/\sqrt(s_1^2/n_1+s_2^2/n_2)= __________________ (Round the answer to 3 decimals)
with degrees of freedom:
df=(s_1^2/n_1+s_2^2/n_2)^2/(((s_1^2/n_1)^2)/(n_1-1)+((s_2^2/n_2)^2)/(n_2-1))

Answer 1

To compute the test statistic \( t_0 \) and the degrees of freedom \( df \), we will use the provided formulas step by step.

### Given Data
- \(\bar{x}_1 = 11.7\)
- \(s_1 = 2.5\)
- \(n_1 = 93\)
- \(\bar{x}_2 = 10.3\)
- \(s_2 = 2.8\)
- \(n_2 = 66\)

### Step 1: Calculate \( t_0 \)

The formula for the test statistic \( t_0 \) is:

\[
t_0 = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}
\]

1. Calculate \( s_1^2 \) and \( s_2^2 \):

\[
s_1^2 = (2.5)^2 = 6.25
\]
\[
s_2^2 = (2.8)^2 = 7.84
\]

2. Calculate \( \frac{s_1^2}{n_1} \) and \( \frac{s_2^2}{n_2} \):

\[
\frac{s_1^2}{n_1} = \frac{6.25}{93} \approx 0.06709677419354838
\]
\[
\frac{s_2^2}{n_2} = \frac{7.84}{66} \approx 0.11878787878787878
\]

3. Add these two results:

\[
\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \approx 0.06709677419354838 + 0.11878787878787878 \approx 0.18588465298142716
\]

4. Take the square root:

\[
\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \approx \sqrt{0.18588465298142716} \approx 0.431777632384792
\]

5. Now compute \( t_0 \):

\[
t_0 = \frac{11.7 - 10.3}{0.431777632384792} = \frac{1.4}{0.431777632384792} \approx 3.239766600560528
\]

Rounding to three decimal places:

\[
t_0 \approx 3.240
\]

### Step 2: Calculate Degrees of Freedom \( df \)

The formula for the degrees of freedom \( df \) is:

\[
df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\left( \frac{\frac{s_1^2}{n_1}^2}{n_1-1} + \frac{\frac{s_2^2}{n_2}^2}{n_2-1} \right)}
\]

1. Calculate \( \left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2 \):

\[
\left( 0.18588465298142716 \right)^2 \approx 0.03463869153212529
\]

2. Calculate the components of the denominator:

- For \( \frac{\frac{s_1^2}{n_1}^2}{n_1-1} \):

\[
\frac{\left(0.06709677419354838\right)^2}{93-1} \approx \frac{0.004493016666205162}{92} \approx 0.00004886697022208552
\]

- For \( \frac{\frac{s_2^2}{n_2}^2}{n_2-1} \):

\[
\frac{\left(0.11878787878787878\right)^2}{66-1} \approx \frac{0.014136095800017755}{65} \approx 0.00021753085384758445
\]

3. Combine these results to compute the denominator:

\[
\left( \frac{\frac{s_1^2}{n_1}^2}{n_1-1} + \frac{\frac{s_2^2}{n_2}^2}{n_2-1} \right) \approx 0.00004886697022208552 + 0.00021753085384758445 \approx 0.000266397824069670
\]

4. Now calculate \( df \):

\[
df \approx \frac{0.03463869153212529}{0.000266397824069670} \approx 130.052
\]

Rounding down, we find:

\[
df \approx 130
\]

### Final Results
- The computed test statistic \( t_0 \) is \( \boxed{3.240} \).
- The degrees of freedom \( df \) is \( \boxed{130} \).