Question

The actual sample data analysis showed that 50% of sample 50 men over the age of 30. Is this sample data different from population data showing 40% of men over the age of 30 suffering from lower back pain? Show me the process how to prove true or false on this statement (use 95% CI).

Answer 1

Solution:

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

n = 50

P = x/n = 50% = 0.50

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.50 ± 1.96* sqrt(0.50*(1 – 0.50)/50)

Confidence Interval = 0.50 ± 1.96* 0.0707

Confidence Interval = 0.50 ± 0.1386

Lower limit = 0.50 - 0.1386 = 0.3614

Upper limit = 0.50 + 0.1386 =0.6386

Confidence interval = (0.3614, 0.6386)

The population proportion of 40% or 0.40 is lies between above confidence interval, so this sample data is not different from the population data.

It is true that the sample data is not different from the population data.

It is false that the sample data is different from the population data.