Question

In a study conducted in the United Kingdom about sleeping positions, 1000 adults in the UK were asked their starting position when they fall asleep at night. The most common answer was the fetal position (on the side, with legs pulled up), with 41% of the participants saying they start in this position. Use a normal distribution to find a 95% confidence interval for the proportion of all UK adults who start sleep in this position. Use the fact that the standard error of the estimate is 0.016.

Answer 1

Answer: = ( 0.411, 0.409)

Therefore at 95% confidence interval (a,b) = (0.411, 0.409)

Explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean gain x = 0.41

Standard deviation r = 0.016

Number of samples n = 1000

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values we have;

0.41+/-1.96(0.016/√1000)

0.41+/-1.96(0.000506)

0.41+/-0.00099

0.41+/-0.001

= ( 0.411, 0.409)

Therefore at 95% confidence interval (a,b) = (0.411, 0.409)