Problem formulating a hypothesis test for the mean Suppose that I would like to support the following hypothesis:
h = The average weight of people in New York is 70kg.
If I take h to be the null hypothesis then I can only reject it (I assume that failing to reject h will not support h).
So, how can I formulate a hypothesis test for supporting h?
 A: Extended comment: Maybe you really want a confidence interval for the mean $\mu$ of the population of weights of people (young adults?) in NYC.
For example, for normally distributed weights (technically masses) with $\mu=70$kg, $\sigma=7$kg and a sample of size $n=1000,$ R code set.seed(515); t.test(rnorm(1000, 70, 7))$conf.int gives 95% CI $(69.635, 70.495)$ for $\mu.$ [To the nearest kg, both endpoints round to $70.]$
set.seed(515)
x = rnorm(1000, 70, 7)
t.test(x)$conf.int
[1] 69.63495 70.49523
attr(,"conf.level")
[1] 0.95

summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  47.42   65.27   70.14   70.07   74.86   92.15 

 hist(x, prob=T, col="skyblue2")


A major difficulty with your question may be to specify
what sub-population of New Yorkers you have in mind.
The actual distribution of all adults in the city is likely
an intricate mixture of diverse normal and gamma distributed populations, and there is no reason to believe
that the overall distribution of their weights is well-modeled
as normal.
Addendum: Very roughly, you might decide to use the distribution $\mathsf{Gamma}(\mathrm{shape}=8,\mathrm{rate}=8/70)$ as a model
for the population, obtaining the sample y of size 1000
below. The population mean is $\mu = 70$ and the sample mean is $\bar Y = 69.68.$
set.seed(2022)
y = rgamma(1000, 8, 8/70)
summary(y)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  19.98   52.19   66.53   69.68   84.67  167.10 
hist(y, prob=T, col="skyblue2")


Then a 95% nonparametric bootstrap CI for $\mu,$ based on data y
is $(68.09, 71.14),$ so it seems that the data are
consistent with $\mu = 70.$
a.obs = mean(y)
d = replicate(2000, mean(sample(y,1000,rep=T))-a.obs)
UL = quantile(d, c(.975,.025))
a.obs - UL
   97.5%     2.5% 
68.09183 71.13845

