Over the past several years, the proportion of one-person households has been increasing. The Census Bureau would like to test the hypothesis that the proportion of one-person households exceeds 0.27. A random sample of 125 households found that 43 consisted of one person. To conduct the hypothesis test, what distribution would you use to calculate the critical value and the p-value?

Answer:For this case we can find the critical value with the significance level [tex]\alpha=0.05[/tex] and if we find in the right tail of the z distribution we got:[tex] z_{\alpha}= 1.64[/tex]The statistic is given by:[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  Replacing we got:  [tex]z=\frac{0.344 -0.27}{\sqrt{\frac{0.27(1-0.27)}{125}}}=1.86[/tex]  Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of households with one person is significantly higher than 0.27Step-by-step explanation:We have the following dataset given:[tex] X= 43[/tex] represent the households consisted of one person[tex]n= 125[/tex] represent the sample size[tex] \hat p= \frac{43}{125}= 0.344[/tex] estimated proportion of  households consisted of one personWe want to test the following hypothesis:Null hypothesis: [tex]p \leq 0.27[/tex]Alternative hypothesis: [tex]p>0.27[/tex]And for this case we can find the critical value with the significance level [tex]\alpha=0.05[/tex] and if we find in the right tail of the z distribution we got:[tex] z_{\alpha}= 1.64[/tex]The statistic is given by:[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  Replacing we got:  [tex]z=\frac{0.344 -0.27}{\sqrt{\frac{0.27(1-0.27)}{125}}}=1.86[/tex]  Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of households with one person is significantly higher than 0.27