Lesson 5.2: Inference for Difference in Means from Two Independent Groups

Software Lab 5.2 Solutions

  1. Calculated value: t=\dfrac{0.401}{0.268} = 1.496. Test output value: 1.50.
    jamovi - independent samples t-test - ncbirths
    Figure 1: Independent samples t-test for ncbirths
  2. Calculated value: df = \dfrac{\left( \dfrac{1.43^2}{100}+\dfrac{1.60^2}{50} \right)^2}{\dfrac{1}{99}\left( \dfrac{1.43^2}{100} \right)^2 + \dfrac{1}{49}\left( \dfrac{1.60^2}{50} \right)^2} = 88.9. Test output value: 89.3.
    jamovi - descriptives - north carolina births
    Figure 2: Group descriptives for North Carolina births data
  3. Calculated p-value: 0.1382. Test output value: 0.138.
  4. The p-value is larger than the significance value, 0.05, so we do not reject the null hypothesis. There is insufficient evidence to say there is a difference in average birth weight of newborns from North Carolina mothers who did smoke during pregnancy and newborns from North Carolina mothers who did not smoke during pregnancy.
  5. Calculated value: t=\dfrac{-0.459}{0.101} = -4.545. Test output value: -4.52.
    jamovi - independent samples t-test - bmi & activity
    Figure 3: Independent samples t-test for BMI and activity
  6. Calculated value: df = \dfrac{\left( \dfrac{4.74^2}{8342}+\dfrac{5.52^2}{4022} \right)^2}{\dfrac{1}{8341}\left( \dfrac{4.74^2}{8342} \right)^2 + \dfrac{1}{4021}\left( \dfrac{5.52^2}{4022} \right)^2} = 6964. Test output value: 6959.
    jamovi - descriptives - high schoolers' physical activity
    Figure 4: Group descriptives for high-schooler physical activity
  7. Calculated p-value: 0.000006. Test output value: < 0.001.
  8. The p-value is smaller than the significance value, 0.05, so we reject the null hypothesis in favour of the alternative hypothesis. There is sufficient evidence to say there is a difference in average body mass index of high schoolers who are physically active at least three days a week and high schoolers who are physically active two or fewer days a week.
  9. (\overline{y}_1-\overline{y}_2) \pm t^* \times \sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}} = -0.459 \pm 1.96(0.101) = -0.459 \pm 0.198 = (-0.657, -0.261). The interval in the “Independent Samples T-Test” output is given as (-0.657, -0.260).
    jamovi - independent sample confidence interval - high schoolers' physical activity
    Figure 5: Independent samples confidence interval for high-schooler physical activity
  10. We’re 95% confident the difference in average body mass index of high-schoolers who are physically active at least three days a week and high-schoolers who are physically active two or fewer days a week is between -0.657 and -0.260. Although statistically significant, this difference is probably not of much practical significance.

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