Answer:
Explanation:
To determine if we can reject the claim that there is no difference in the proportion of college enrollees between the two groups (females and males), we need to perform a hypothesis test.
Let:
p1 = Proportion of females enrolled in college
p2 = Proportion of males enrolled in college
We are given the following information:
n1 = Sample size for females = 190
n2 = Sample size for males = 160
x1 = Number of females enrolled in college = 0.70 * 190 = 133
x2 = Number of males enrolled in college = 0.60 * 160 = 96
(a) State the null and alternative hypotheses:
Null hypothesis (H0): p1 - p2 = 0 (There is no difference in the proportion of college enrollees between the two groups)
Alternative hypothesis (Ha): p1 - p2 ≠ 0 (There is a difference in the proportion of college enrollees between the two groups)
(b) Calculate the pooled sample proportion (p):
p = (x1 + x2) / (n1 + n2)
p = (133 + 96) / (190 + 160)
p = 229 / 350
p ≈ 0.6543
(c) Calculate the standard error (SE) of the difference in proportions:
SE = √[p * (1 - p) * (1/n1 + 1/n2)]
SE = √[0.6543 * (1 - 0.6543) * (1/190 + 1/160)]
SE ≈ √[0.6543 * 0.3457 * (0.005263 + 0.00625)]
SE ≈ √[0.227 * 0.011513]
SE ≈ √0.002614
SE ≈ 0.05113
(d) Calculate the test statistic (z-score):
z = (p1 - p2) / SE
z = (0.70 - 0.60) / 0.05113
z ≈ 1.954
(e) Conduct the hypothesis test at α = 0.09:
At α = 0.09 (or 9% significance level), we have a two-tailed test.
The critical z-value for a two-tailed test at α = 0.09 is approximately ±1.695.
Since the calculated z-score (1.954) falls outside the critical region of ±1.695, we can reject the null hypothesis.
Conclusion:
Based on the hypothesis test, we can reject the claim that there is no difference in the proportion of college enrollees between the two groups (females and males) at α = 0.09 significance level. There is evidence to suggest that there is a difference in the proportion of college enrollees between the two groups.