what is difference between $95 \%$ CI of mean and 95% pdf of normal distribution?

Question

We took sample mean $\mu = 14$, and $\sigma = 0.45$.

Calculate required area of normal distribution

We will apply $\int_{13.19}^{15} f(x) \ dx = 95\%$.

95 % CI of mean

But if we took sample size $n = 9$ and $z_{95\%} = 1.96$, hence

$95 \% $ CI = $14 \pm 1.96(\frac{0.45}{\sqrt{9}})$ = $lower \ limit\ (13.7) \ and \ upper \ limit\ (14.2940)$.

My question, If we apply these lower and upper limit in pdf function, then $\int_{13.7}^{14.2940} f(x) \ dx =49 \%$.

We are excited to know, how we can make difference $95 \%$ area of pdf function and $95 \%$ CI of mean.

Answer 1

You have a normal distribution, $N(\mu,\sigma^2)$, where $\mu$ is unknown and $\sigma = 0.45$. The central 95% of the distribution’s area is between $\mu – 1.96\sigma$ and $\mu + 1.96\sigma$. You took $N = 9$ iid samples from the distribution and computed their sample mean, $m$, as $14$. The sample mean is a sample from the sampling distribution of the mean, i.e., a sample from $N(\mu,\sigma^2/N)$, where $N = 9$. Thus the central 95% confidence interval for $\mu$ is $m ± 1.96\sigma/√9$. This is $14 ± 1.96(0.45/3)$, i.e., from 13.706 to 14.294. So you have two distributions: $N(\mu,\sigma^2)$ (from which you drew the $N = 9$ iid samples) and $N(\mu,\sigma^2/N)$ (from which m is a sample).

The interval $\mu ± 1.96\sigma/√9$, is only 48.6% of the area of $N(\mu,\sigma^2)$, but this is irrelevant: the interval applies to $N(\mu,\sigma^2/9)$.