Question

The Bureau of Labor Statistics (BLS) reported that 9.9% of adults aged 20 to 24 years were unemployed in February 2015. Suppose that Jacqui believes that in her city the unemployment rate is higher than the reported proportion. She determines the unemployment status of 450 randomly selected adults in the given age range. She decides to conduct a z-test for the proportion p of unemployed adults in her town aged 20 to 24 years old. Determine the power to reject the null hypothesis at a significance level of a = 0.05 if the true proportion of unemployed adults aged 20 to 24 is 0.110. Use software if you have access to software that computes power, otherwise do the computation by hand. Give your answer to two decimal places, rounding if necessary. power :

Answer 1

As we are testing here whether the population proportion is more than 0.099, therefore this is a one tailed test here. For 0.05 level of significance, we have from standard normal tables here as:
P(Z > 1.645) = 0.05

The critical proportion value here is computed as:

$\frac{p_{crit} - P}{\sqrt{\frac{P(1-P)}{n}}} = 1.645$

$p_{crit} = P + 1.645\sqrt{\frac{P(1-P)}{n}} = 0.099 + 1.645*\sqrt{\frac{0.099(1 - 0.099)}{450}}$

For a true proportion value of P = 0.110, the power of the test is computed as the probability of rejecting the null hypothesis which is computed here as:

P( p > 0.1081)

Converting it to a standard normal variable, we have here:

$P(Z > \frac{0.1081 - 0.11}{\sqrt{\frac{0.11*0.89}{450}}})$

Getting it from the standard normal tables, we have here:

Therefore 55.17% is the required power of the test here.