= Solution
Let $\mu_0,\mu_1$ be the <population means> of the numerical <vaccination intention> score in the two recorded groups. The <two-sided hypothesis test> compares
$$
H_0:\mu_0-\mu_1=0,\qquad H_1:\mu_0-\mu_1\ne0.
$$
The <sample means> are $13.48553$ and $13.46565$, giving a <difference of sample means> of approximately \b[0.01988 score points], in the order group zero minus group one. The reported group counts are 1,866 and 262; seven observations have a missing group indicator.
<Welch's t-test> divides that estimated difference by its estimated <standard error>. Its reported <test statistic> is $t=0.055032$, with approximately $344.43$ <statistical degrees of freedom>. This is an unequal-variance comparison; fractional <statistical degrees of freedom> are appropriate for the Welch approximation. The two groups need not have equal <sample sizes> or equal population <variances>.<Two-Sample t-Test for Equal Means by National Institute of Standards and Technology>[Section 1.3.5.3, unpaired test statistic, unequal-variance standard error, and Welch–Satterthwaite degrees of freedom]{ref}
The reported <p-value> is $0.9561$. Under the specified equal-mean <null hypothesis> and sampling model, a <p-value> measures the probability of a result at least as extreme as the observed one. It is not a posterior probability that the <null hypothesis> is true. \b[There is no rejection at a prespecified 5% significance level.] The observed <difference of sample means> is tiny relative to its estimated <standard error>.
The 95% <confidence interval> is approximately
$$
\boxed{\mu_0-\mu_1:\quad[-0.69071,\ 0.73047].}
$$
It includes zero and differences in either direction. The interval's coverage refers to repeated use of the procedure under its assumptions, rather than to a probability assigned to this fixed unknown parameter or to the spread of individual scores. Nonrejection does not establish that the two <population means> are identical.
<Simple random sampling> supports the sampling interpretation, but the score is bounded and its raw distribution need not be a <normal distribution>. The large group sizes can support an approximate normal sampling distribution of the <sample means> under regularity conditions. The supplied <observational study> does not identify a causal explanation of any difference, and the numeric coding of the group labels creates no biological or ordinal interpretation.
Back to article page
