= Solution
\b[TODO: Attachment C is absent from the held release. Recover it to check any supplied critical-value table, significance convention or additional instructions. This leaf remains incomplete against that missing input.] The printed summary statistics nevertheless allow the following conditional numerical answer, using a two-sided $5\%$ <Student's t-test> and independent observations with common <variance>.
Test $H_0:\mu_W=\mu_S$ against $H_1:\mu_W\ne\mu_S$. The pooled <sample variance> is
$$
s_p^2=\frac{(1430-1)(3.29)^2+(1569-1)(3.22)^2}{1430+1569-2}=10.5856824.
$$
The <test statistic> and <statistical degrees of freedom> are
$$
t=\frac{5.72-5.36}{\sqrt{10.5856824(1/1430+1/1569)}}=3.02646,\qquad \nu=2997.
$$
Using the <Student t-distribution> directly gives a two-sided <p-value> $p\simeq0.00250$ and a $5\%$ critical value $1.96076$. Thus \b[reject equal seasonal means at $5\%$ and $1\%$; the winter mean is higher by $0.36$ concentration units]. This does not establish a large physical effect or its cause.
The two <variances> are similar, and <Welch's t-test> gives essentially the same result: $t=3.02344$, $\nu\simeq2958.36$, $p\simeq0.00252$. Both calculations assume the counts represent independent observations. Serial dependence, repeated measurements and strong non-normality can invalidate nominal <p-values>; the summary table alone cannot diagnose them. The unequal means are a conclusion under these statistical assumptions, pending recovery of Attachment C.
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