Question

The following data represent the age (in weeks) at which 10 randomly selected babies crawled for the first time. 43, 44, 42, 31, 38, 45, 36, 39, 50, 52 Suppose that the data is a random sample from a population with a normal distribution. Construct a 90 confidence interval for the average age at which babies first crawl.

Answer 1

Answer:

$42-1.833\frac{6.325}{\sqrt{10}}=38.334$    

$42+1.833\frac{6.325}{\sqrt{10}}=45.666$    

The 90% confidence interval is given by $38.334 \leq \mu \leq 45.666$

Step-by-step explanation:

Notation

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)  

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)  

The mean calculated for this case is $\bar X=42$

The sample deviation calculated $s=6.325$

The degrees of freedom are given by:

$df=n-1=10-1=9$

The Confidence level is 0.90 or 90%, the significance is $\alpha=0.1$ and $\alpha/2 =0.05$, and the critical value would be $t_{\alpha/2}=1.833$

Replacing we got:

$42-1.833\frac{6.325}{\sqrt{10}}=38.334$    

$42+1.833\frac{6.325}{\sqrt{10}}=45.666$    

The 90% confidence interval is given by $38.334 \leq \mu \leq 45.666$