Additionally, could you break it down step by step. Thank you.

 

A high school that is interested in change course scheduling for seniors to help them prepare for college believed that if they had a free period they would use it in to morning to start classes later in the day. To test this and see if it would hold true the incoming senior class and would reach out to some alumni in there first or second year of college. Out of the senior class only 40 out of 125 would chose to start later in the day, while the individuals in college 80-139 said they start their schedules latter in the day if given the chance.

hypothesis test with the given data using a .05 level of significance.

Let p₁= population proportion of high school(senior class) would chose to start later in the day

p₂=population proportion college said they start their schedules latter in the day if given the chance

p₁=40/125=0.32, n₁=125

p₂=80/139=0.58, n₂=139

Null Hypothesis:

Ho: p₁ = p₂

Alternative Hypothesis:

Ha: p₁ ≠ p₂

Test statistic for two population proportion:

?p^?=(125+139)(40+80)?=0.45?

?z=p^?(1−p^?)(n1?1?+n2?1?)?p1?−p2??=0.45(1−0.45)(1251?+1391?)?0.32−0.58??

?z=−4.24?

Critical Value by two-tailed test and using Excel Function:

α/2=0.05/2=0.025

Excel Formula:

=NORM.S.INV(0.025)

Z_c=±1.96

Since the test statistic(-4.24) which is negative is less than the negative value of critical value(-1.96), we reject null Hypothesis Ho.

Therefore, there is enough evidence that the incoming senior class and would reach out to some alumni in there first or second year of college is true.