Answer:
Given:
n = 550
Po = 51% = 0.51
P0 - 1 = 0.51 - 1 = 0.49
p' = [tex] \frac{287}{550} [/tex] = 0.5218
np0(1 - p0) = 550*0.51(1-0.51)
= 137.4 ≥ 100
This means the sample size is greater than 5% of the population size. and the sample can be reasonably assumed to be random, the requirement for testing hypothesis are satisfied.
1) The null and alternative hypotheses:
H0 : p0 = 0.51
H1 : p0 ≠ 0.51
The test statistic Z,
[tex] Z = \frac{p' - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} [/tex]
[tex] Z = \frac{0.5218 - 0.51}{\sqrt{\frac{0.51(1-0.51)}{550}}} [/tex]
Z = 0.55
This is two tailed test.
The pvalue for Z =0.55
Pvalue = 0.5824
3) Decision:
Since pvalue, 0.5824 is greater than significance level, we fail to reject null hypothesis H0.
Conclusion:
Do not reject Upper H0. There is not sufficient evidence at the alpha equals 0.01 level of significance to conclude that the proportion of females who are living alone has changed.