Question

A health research company was engaged to test the accuracy of COVID-19 test kits used in Queensland. A sample of 50 test kits was analysed to see whether they gave the correct test result. An interval estimate of 0.46 to 0.70 was found for the population proportion of test kits which gives the correct test result. What is the level of confidence used for this test? (Answer as a percentage to two decimal places. Do not include the % symbol).

Answer 1

Sol:

lower limit=0.46,sample prop-margin of error=0.46

upper limit=0.70,sample prop+margin of error=0.70

Adding 2 eq

2p^=0.46+0.70

p^=(0.46+0.70)/2

=0.58

margin of error=upperlimit-sample proportion

=0.70-0.58

= 0.12

standard error

sqrt(p^*(1-p^)/n)

=sqrt(0.58*(1-0.58)/50)

= 0.06979971

margin of error=zcrti*std error

0.12=zcrit*0.06979971

zcrit=0.12/0.06979971

=1.719205

=1.72

area corresponding to z=1.72 is 0.95728

alpha=level of significance=1-0.95728= 0.04272=0.04272*100

= 4.272

=4.27%