Question

Construct a $95\%$ confidence interval for the population mean weight of newborn elephants. State the confidence interval, sketch the graph and calculate the error bound.

Short answer

We estimate with $95\%$ confidence that the true population for weight of newborn elephants is between $239.84$ and $248.16$.

Step-by-step solution

Given Information

Fifty newborn elephants are weighed. The sample mean is $244$ pounds. The sample standard deviation is $11$ pounds.

Explanation 

If $\overline{x}$is the sample mean of a random sample of size $n$from a normal population with unknown variance ${\sigma }^2$,

$\overline{x}-z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}\le \mu \le \overline{x}+z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$

where $z_{\frac{\alpha }{2}}$is the upper $100\frac{\alpha }{2}$percentage point of the standard normal distribution.

The standard deviation of the weights of elephants is known to be approximately $\sigma =15$pounds. A random sample size $n=50$and a sample mean $\overline{x}=244$pounds.

We need find a $95\%$ confidence interval estimate for the population mean weight of newborn elephants. Therefore,

$\frac{\alpha }{2}=\frac{1-0.95}{2}=0.025\Rightarrow z_{\frac{\alpha }{2}}=z_{0.025}=1.96$

Explanation

The previous implication was obtained on a probability table for the standard normal distribution.

From (1) and (2) we get

$244-1.96\frac{15}{\sqrt{50}}\le \mu \le 244+1.96\frac{15}{\sqrt{50}}$

Therefore, $95\%\mathrm{Cl}$ for $\mu$ is

$239.84\le \mu \le 248.16$

We estimate with $95\%$confidence that the true population for weight of newborn elephants is between $239.84$and $248.16$

Ninety percent of all confidence intervals constructed in this way contain the true mean statistics exam score. For example, if we constructed $100$of these confidence intervals, we would expect $95$of them to contain the true population mean.

Final Answer

Confidence interval $CI=(239.84,248.16)$

The graph:

Error bpund$EBM:4.16$