Question

A human resources specialist is investigating a government claim that the unemployment rate is less than 5%. To test this claim, a random sample of 1500 people is taken and its determined that 61 people are unemployed. The following is the setup for this hypothesis test:

H0:p=0.05

Ha:p<0.05

Find the p-value for this hypothesis test for a proportion and round your answer to 3 decimal places.

The following table can be utilized which provides areas under the Standard Normal Curve:

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
-1.8 0.036 0.035 0.034 0.034 0.033 0.032 0.031 0.031 0.030 0.029
-1.7 0.045 0.044 0.043 0.042 0.041 0.040 0.039 0.038 0.038 0.037
-1.6 0.055 0.054 0.053 0.052 0.051 0.049 0.048 0.047 0.046 0.046
-1.5 0.067 0.066 0.064 0.063 0.062 0.061 0.059 0.058 0.057 0.056
-1.4 0.081 0.079 0.078 0.076 0.075 0.074 0.072 0.071 0.069 0.068

Answer 1

P-value =0.062

At a signficance level of 0.05, there is not enough evidence to support the claim that the unemployment rate is less than 5%.

Explanation:

This is a hypothesis test for a proportion.

The claim is that the unemployment rate is less than 5%.

Then, the null and alternative hypothesis are:

`H_0: \pi=0.05\\\\H_a:\pi<0.05`

The significance level is 0.05.

The sample has a size n=1500.

The sample proportion is p=0.041.

`p=X/n=61/1500=0.041`

The standard error of the proportion is:

`\sigma_p=\sqrt{(\pi(1-\pi))/(n)}=\sqrt{(0.05*0.95)/(1500)}\\\\\\ \sigma_p=√(0.000032)=0.0056`

Then, we can calculate the z-statistic as:

`z=(p-\pi+0.5/n)/(\sigma_p)=(0.041-0.05+0.5/1500)/(0.0056)=(-0.0087)/(0.0056)=-1.5401`

This test is a left-tailed test, so the P-value for this test is calculated as:

`\text{P-value}=P(z<-1.5401)=0.062`

As the P-value (0.062) is greater than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the unemployment rate is less than 5%.