Question

A study of 40 SHS English teachers established that they spent 12.6 minutes revising a student’s term paper.

Find the 90% confidence interval in the mean time for all composition papers when minutes.

If the teacher affirmed that she spent an average of 30 minutes revising a term paper, what would be your reaction?

 

Find the sample size to estimate a population mean within 95% confidence if the standard deviation is 6.2.

 

The prices for LED are listed: 22 500, 24 000, 21 500, 20 200, 20 600, 21 100, 210 000, 19 300, 25 000 and 22 500. Estimate the true mean of the data set with 90% confidence level.

 

Louie wants to estimate the average value of the lot in his town with 99% confidence interval. Use his random sample of 36 lots with an average value of Php251, 131.42 and a standard deviation of Php1 321.467 to find the confidence interval.

Answer 1

1. $CI=\mu \plusmn z \frac{\sigma}{\sqrt{n}}$

Z(90%)=1.64

n=40

$\mu=12.6$

$CI=12.6 \plusmn 1.64 \frac{\sigma}{\sqrt{40}}=12.6 \plusmn 0.26 \sigma$

$30=12.6+0.26 \sigma$

$\sigma=(30-12.6)/0.26=66.9$

If σ smaller then 66.9 min, then 30 min is too long.

If σ larger then 66.9, then it is ok.

2. $n=(\frac{z\sigma}{0.5})^2=(\frac{1.96x6.2}{0.5})^2=591$

0.5 is a measure of accuracy. Usually it equal 50%=0.5 , if other is not written in the task.

3. m=(22500+24000+21500+20200+20600+21100+210000+19300+25000+22500)/10=40670

$\sigma=\sqrt{\frac{\sum (x-m)^2}{n-1}}=\sqrt{\frac{330148900+377888900+367488900+419020900+402804900+382984900+2.867x10^{10}[phone]+245548900+330148900}{9}}=59833$

$CI=m \plusmn z \frac{\sigma}{\sqrt{n}}=40670 \plusmn 1.645 \frac{59833}{\sqrt{10}}=40670 \plusmn 31124.8$

4.$CI=\mu \plusmn z\frac{\sigma}{\sqrt{n}}=251,131.42 \plusmn 2.58 \frac{1321.467}{\sqrt{36}}=251,131.42 \plusmn 568.23$