Let p represents the population proportion.
Then According to the given information, we have
[tex]H_0: p=0.125\\\\H_a: p>0.125[/tex]
∵ the alternative hypothesis is right-tailed , so the test is right-tailed test.
Given : For sample size of n=410 U.S. workers is collected in 2006 , 50 of the workers belonged to unions.
Then , sample proportion : [tex]\hat{p}=\dfrac{50}{410}\approx0.12[/tex] [Rounded to 2 decimals]
Test statistic: [tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
[tex]z=\dfrac{0.12-0.125}{\sqrt{\dfrac{0.125(1-0.125)}{410}}}\\\\=-0.306127891108\approx-0.31[/tex] [Rounded to 2 decimals]
The value of the test statistic : z= -0.31
P-value (Right -tailed test)=[tex]P(z>-0.31)=P(z<0.31)=0.6217195\approx0.6217[/tex] [Rounded to 4 decimals]
Since , the p-value (0.6217) is greater than the significance level (0.05), so we accept the null hypothesis.
Conclusion: We have sufficient evidence to reject the alternative hypothesis that the union membership increased in 2006.