Question

a sample of the length in inches for newborns is given below. assume that lengths are normally distributed. find the 95% confidence interval of the mean length. do not round in between steps. round answers to at least 4 decimal places. question help: video message instructor

Explanation:

Step1: Calculate the sample mean $\bar{x}$

The formula for the sample mean is $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$.
Here, $n = 10$, and $\sum_{i=1}^{10}x_{i}=20.2 + 15.6+15.6 + 18.7+16.1+17.8+15.7+20.3+15.1+20.2=175.3$.
So, $\bar{x}=\frac{175.3}{10}=17.53$.

Step2: Calculate the sample standard deviation $s$

The formula for the sample standard deviation is $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$.
First, calculate $(x_{i}-\bar{x})^{2}$ for each $x_{i}$:
$(20.2 - 17.53)^{2}=7.1289$, $(15.6-17.53)^{2}=3.7249$, $(15.6 - 17.53)^{2}=3.7249$, $(18.7-17.53)^{2}=1.3689$, $(16.1-17.53)^{2}=2.0449$, $(17.8-17.53)^{2}=0.0729$, $(15.7-17.53)^{2}=3.3489$, $(20.3-17.53)^{2}=7.6729$, $(15.1-17.53)^{2}=5.9049$, $(20.2-17.53)^{2}=7.1289$.
$\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=7.1289+3.7249+3.7249+1.3689+2.0449+0.0729+3.3489+7.6729+5.9049+7.1289 = 42.121$.
Then, $s=\sqrt{\frac{42.121}{9}}\approx2.1607$.

Step3: Determine the critical value $t_{\alpha/2}$

Since the confidence level is $95\%$, $\alpha=1 - 0.95 = 0.05$, and $\alpha/2=0.025$. The degrees of freedom $df=n - 1=9$.
From the $t$-distribution table, $t_{0.025,9}=2.2622$.

Step4: Calculate the margin of error $E$

The formula for the margin of error is $E = t_{\alpha/2}\frac{s}{\sqrt{n}}$.
Substitute the values: $E=2.2622\times\frac{2.1607}{\sqrt{10}}\approx2.2622\times0.6833\approx1.5467$.

Step5: Calculate the confidence interval

The confidence interval is $\bar{x}-E<\mu<\bar{x}+E$.
Substitute the values: $17.53-1.5467 <\mu<17.53 + 1.5467$.

Answer:

$15.9833<\mu<19.0767$