Question

A recent year was randomly picked from 1985 to the present. In that year, there were 2051 Hispanic students at Cabrillo College out of a total of 12,328 students. At Lake Tahoe College, there were 321 Hispanic students out of a total of 2441 students. Construct a 90% confidence interval for the difference.

What conclusion can you draw?

Answer 1

Solution :

P1 2051 12328 0.166

321 P2 = 0.132 2441

Pooled sample proportion (P),

P-Pini + P2N2 0.166 * 12328 + 0.132 * 2441 12328 + 2441 - 0.16 ni + n2

At 90% confidence inerval the critical value of z_0.05 = 1.645

The 90% Confidence interval is given by,

1 (pa – p) #zy/P(1 – P) (m+ 12

1 1 (0.166 – 0.132) = 1.645/0.16 * 0.84 * + 12328 2441

0.034 +1.645 +0.0081

0034-00133

(0.0207.0.0473)

At 90% Confidence Interval for the difference is (0.0207,0.0473).

We can say that there is only a 10% chance that the range 0.0207 to 0.0473  excludes the difference between the two group.

Also, if the 90% CI does not contain the value 0, then the p-value will be strictly less than 0.1. Hence we can reject the null hypothesis.

So if p₁ = Hispanic students at Cabrillo College and

p₂ = Hispanic students at Lake Tahoe College

then the null and alternate hypothesis will be,

Ho: P1 = P2

ad id: "H

As the confidence interval doesn't contain 0, hence p-value is less than 0.1. Therefore we reject the null hypothesis.

Hence, Hispanic students at the two college is different.

**Thank you for giving me the opportunity to help you. Please give me thumbs up if you like this answer.**