 Solution

There are 14 conservative-coded and 12 liberal-coded observations. The tweet summary has sample median 258.5, quartiles 204.5 and 323.2, and range 108 to 441, so the interquartile range is about 118.7 tweets. No measurement period is specified, so these should not be reinterpreted as tweets per day. The displayed overall sample mean is 261.5, subject to a further consistency issue below. These are small groups; the summary alone cannot establish normality, lack of outliers or an appropriate sampling design.

Let ${\mu }_C$ and ${\mu }_L$ be the two population means for the defined tweet measure. The null hypothesis is $H_0:{\mu }_C-{\mu }_L=0$, against the two-sided test alternative $H_1:{\mu }_C-{\mu }_L\ne 0$. Welch's t-test usesallowing unequal group variances. It requires independent observations and a suitable approximation to the sampling distribution of the mean difference; a large-sample justification cannot simply be assumed with group sizes 14 and 12. Its noninteger statistical degrees of freedom comes from the Welch approximation, not a fractional number of respondents.^[ref]

The two displayed sample means implyThe negative sign means the liberal-coded group has the larger observed mean. This is a difference in group averages, not a statement about every individual and not evidence that political orientation causes tweeting behaviour. Practical importance concerns this difference and its context; statistical significance does not itself measure its importance.

A p-value is calculated under the null hypothesis and the assumed sampling model. It is the probability of a result at least as incompatible with that hypothesis as the observed one, not the probability that the hypothesis is true or that this conclusion is mistaken. A prespecified significance level controls the long-run Type I error rate of rejecting a true null; a Type II error is nonrejection of a false null, and statistical power concerns rejection under a specified alternative. Neither a small p-value nor a nonrejection settles the effect size or a causal explanation.

Taken on its own, the printed p-value supports a nominal 5% rejection, but not a 1% rejection. Taken instead from the printed t statistic and statistical degrees of freedom, the two-sided tail is about 0.06380. These different 5% decisions cannot be reconciled by ordinary rounding. Conditional on trusting the group means, t and degrees of freedom, the implied standard error is $(-13.4881)/(-1.9436)\approx 6.93975$ and the corresponding 95% confidence interval isIt includes zero, consistently with that larger p-value. This calculation is a conditional consistency check, not recovery of the true raw-data interval.

The source lists its interval endpoints as 0.04582 and $-126.7904$, in that order. They are not ordered, and merely sorting them would still give midpoint about $-63.3723$, rather than the estimated difference $-13.4881$. The symmetric confidence-interval consistency identities therefore fail for the usual t interval. A frequentist 95% interval describes repeated-sample coverage of its procedure, not a 95% posterior probability for a fixed parameter; neither interpretation justifies silently repairing these endpoints.

The group summaries also imply pooled mean $(14\times 255.4286+12\times 268.9167)/26\approx 261.6539$, which cannot round to the printed 261.5. The pooled-mean compatibility under rounding identifies this additional mismatch for summaries of the same complete sample. Thus the observed direction is conditionally liberal-greater-than-conservative, but the inferential decision and interval require corrected coherent output. Nonrejection under the t/df reading would not establish equal population means. Random assignment, representativeness and possible confounding are not supplied by this test.