Question

Construct a confidence interval for p based on the sample data given here: 12.3 11.6 11.9 12.8 11.6 11.4 12 11.7 11.8 12.6 Use 83% as the confidence level. Round the final values to two digits after the decimal point. The lower limit in the interval is: Number and the upper limit in the interval is: Number

Answer 1

From the given sample data :

The sample size is , n=10

The sample mean is , $\bar{X}=\frac{\sum X}{n}=11.97$

The sample standard deviation is , $s=\sqrt{\frac{\sum X^2-\frac{(\sum X)^2}{n}}{n-1}}=0.4596$

The significance level is , $\alpha=0.17$

Now, df=degrees of freedom=n-1=10-1=9

The critical value is , $t_{df,\alpha/2}=t_{9,0.17/2}=1.492$ ; The Excel function is , =TINV(0.17,9)

Therefore , the 83% confidence interval is ,

$\bar{X}\pm t_{df,\alpha/2}\frac{s}{\sqrt{n}}$

$11.95\pm 1.492*\frac{0.4596}{\sqrt{10}}$

$11.95\pm 0.2168$

Therefore , the lower limit=11.75 and upper limit=12.19