Question

Statistical Significance. In an experiment testing a method of gender selection intended to increase the likelihood that a baby is a girl, 1000 couples give birth to 501 girls and 499 boys. An analyst argues that this fails to provide evidence that the method has any effect, because the probability of getting these results by chance is 0.487, or almost 50%.

Answer 1

Null hypothesis Ho : P = 0.5

Alternate hypothesis Ha : P > 0.5

As if likelihood has increased then, proportion would be greater than 0.5

N = 1000

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 500

N*(1-p) = 500

Both the conditions are met so we can use standard normal z table to estimate the P-Value

Test statistics z = (oberved p - claimed p)/standard error

Standard error = √{claimed p*(1-claimed p)/√n

Observed P = 501/1000

Claimed P = 0.5

N = 1000

After substitution

Z = 0.06

From z table, P(z>0.06) = 0.4761

P-value = 0.4761

As the p-value is extremely large we fail to reject the null hypothesis

So we do not have enough evidence to conclude that

method of gender selection intended to increase the likelihood that a baby is a girl