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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))

1 answer

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 \( \boxed{130} \).