Compare Two Independent Means

Birth Weights and Smoking

• Do infants whose mothers smoke have a different mean birth weight than infants whose mothers do not smoke?
• Let \(\mu_n\) be the mean weight (lbs) of infants whose mothers did not smoke, and let \(\mu_s\) be the mean for infants whose mothers smoked

Inference

• We will estimate the difference in mean birth weights \(\mu_n-\mu_s\) using a confidence interval
• We will conduct a hypothesis test with hypotheses
  • \(H_0: \mu_n-\mu_s = 0\)
  • \(H_A: \mu_n-\mu_s \neq 0\)

Data

• The births14 dataset is available here
• Random sample of 1,000 cases from US birth data set from 2014 (19 removed with missing values)
• habit is smoking habit (“smoker” or nonsmoker”)
• weight is birth weight in pounds

EDA

Randomization Test for Difference in Means

• We can simulate a true null hypothesis by randomly permuting the values of the response variable
• This simulates samples in which there is no association between habit and weight since the values of weight are randomly assigned to the cases
• We can use the Randomize module in Jamovi to do this

Histogram of differences in means (null distribution) calculated from 1,000 random permutations of birth weights. Observed difference is indicated by dashed vertical line.

• There are 0 simulated differences that are \(\geq 0.59\)
• There are 1,000 simulated differences that are \(\leq 0.59\)
• p-value = \(2\times\frac{0}{1000} = 0\)

Test Statistic for Comparing Two Means

• The test statistic for comparing two means is the \(T\) statistic (\(T\) score)
• We will use a version of the \(T\) statistic that assumes the two populations have equal variance (different than the version presented in the text)

Pooled Sample Standard Deviation

• First we compute the pooled sample standard deviation, \[s_p = \sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}\]
• The pooled sample standard deviation in birth weights is \[\begin{array}{rcl} s_p &=& \sqrt{\frac{(867-1)\cdot 1.23^2+(114-1)\cdot 1.60^2}{867+114-2}}\\ &=& 1.28\end{array}\]

• The \(T\) statistic is \[T=\frac{(\bar{x}_1-\bar{x}_2)-0}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\]
• For the birth weight example, the value is \[T=\frac{0.59-0}{1.28\cdot\sqrt{\frac{1}{867}+\frac{1}{114}}} = 4.63\]

Mathematical Model for Testing the Difference in Means

Two Sample T-Test

• The degrees of freedom for the birth weight example is \(df=867+114-2=979\).
• We can use the Randomize module in Jamovi to calculate a p-value using a T distribution with 979 degrees of freedom
• Find the area that is above 4.63 (the observed value of \(T\)) or below -4.63

The \(t\)-distribution with 979 degrees of freedom. The observed \(T\)-statistic is 4.63. The p-value is the total area to the left of -4.63 or to the right of 4.63 (too small to be visible).

• The p-value is extremely small, p-value < 0.001

Note About Relaxing the Equal Variance Assumption

• We can compute a \(T\) statistic without assuming equal variances (see the formula in the book)
• If the null hypothesis is true and the technical conditions are met, then the distribution of these \(T\) statistics will be be approximately \(t\)-distributed
• The \(df\) for the approximating \(t\) distribution involves a complicated calculation (not the one listed in the text)

• We can use the Independent Samples T-Test module in Jamovi to calculate a p-value
• It calculates the \(T\) statistic, \(df\), and the p-value for us
• It does NOT check conditions
• Student’s t-test assumes equal variances, Welch’s t-test does not

Bootstrap Confidence Interval for Difference in Means

• The method for calculating bootstrap confidence intervals for a difference in means is similar to the methods we used for a difference in proportions
• We create bootstrap samples from each group (resampling with replacement) and calculate a difference in means
• We do this 1,000 times and use the resulting sampling distribution to calculate a bootstrap percentile CI
• We can use the Randomize module in Jamovi to do this

1,000 bootstrapped means with dashed lines at 2.5% and 97.5% percentiles.

• The 95% bootstrap percentile CI is (0.312, 0.917)

Estimating the Difference in Means Using a Mathematical Model

• If the technical conditions are met, including the equal variance assumption, then we can use the \(t\)-distribution to estimate the difference in means
• We can calculate a confidence interval for the difference in means as \[(\bar{x}_1-\bar{x}_2)\pm t^{\ast}_{df}\times SE\]

• Assuming equal variance, \(df=n_1+n_2-2\), and the standard error is \[SE = s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\]
• The value of \(t^{\ast}_{df}\) depends on the degrees of freedom and the confidence level
• For the birth weights example \[SE=1.28\cdot\sqrt{\frac{1}{867}+\frac{1}{114}}=0.128\]

• We can use the Randomize module in Jamovi to calculate the appropriate value of the multiplier \(t^{\ast}_{df}\) for a desired confidence level and degrees of freedom
• Since \(df=979\) for this example, the value of \(t^{\ast}_{df}\) for a 95% confidence interval is 1.96
• Thus, the 95% confidence interval is \[0.59\pm1.96\times0.128=0.59\pm0.251\]
• In interval form it is approximately (0.339, 0.841)

Relaxing the Equal Variance Assumption for CI

• We can also use the Independent Samples T-Test analysis in Jamovi to calculate a CI for the mean difference
• We can assume equal variances (Student’s) or not (Welch’s)

Conclusions

• We get the same result from both versions of the hypothesis test.
• We reject the null hypothesis at the \(\alpha=0.05\) significance level and conclude that there is strong evidence of a difference in the average weights of infants born to mothers who smoked and those who did not (p < 0.001)

• We are 95% confident that the mean weight of babies born to mothers who did not smoke is between 0.285 and 0.900 pounds higher than the mean weight of babies whos mothers smoked
• This result is also consistent with the result of the hypothesis test, since 0 does not appear in the 95% confidence interval