Question

The weights at birth of five randomly chosen baby giraffes were 111, 115, 120, 103, and 106 pounds. Assume the distribution of weights is normally distributed. Find a95% confidence interval for the mean weight of all baby giraffes. Use technology for your calculations. Give the confidence interval in the form "estimate + margin oferror." Round to the nearest tenth

Answer 1

We have to calculate a 95% confidence interval for the mean.

The data is: [111, 115, 120, 103, 106].

We can calculate the sample mean as:

$\begin{gathered} M=\dfrac{1}{n}\sum ^n_{i=1}\, x_i \\ M=\dfrac{1}{5}(111+115+120+103+106) \\ M=\dfrac{555}{5} \\ M=111 \end{gathered}$

and the sample standard deviation as:

$\begin{gathered} s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2} \\ s=\sqrt{\dfrac{1}{4}((111-111)^2+(115-111)^2+(120-111)^2+(103-111)^2+(106-111)^2)} \\ s=\sqrt{\dfrac{186}{4}} \\ s=\sqrt{46.5}\approx6.82 \end{gathered}$

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{6.82}{\sqrt{5}}=\dfrac{6.82}{2.236}=3.05$

The degrees of freedom for this sample size are:

$df=n-1=5-1=4$

The t-value for a 95% confidence interval and 4 degrees of freedom is t=2.776.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=2.776\cdot3.05\approx8.5$

Then, we can express the confidence interval as:

$\begin{gathered} CI=M\pm\text{MOE} \\ CI=111\pm8.5 \end{gathered}$

Answer:

The 95% confidence interval for the weights of all baby giraffes is 111 ± 8.5.