In 2011, a U. S. Census report determined that 68% of college students are working students. A researcher thinks this percentage has changed and surveys 144 college students. The researcher reports that 93 of the 144 are working students. Is there evidence to support the researcher's claim at the 1% significance level

We will fail to reject the null hypothesis and conclude that there is sufficient evidence not to support the researchers claim.

Step-by-step explanation:

We are given;

Population proportion; p = 68% = 0.68

Sample proportion; p^ = 93/144 = 0.6548

n = 144

The hypotheses are:

Null hypothesis;H0: p = 0.68

Alternative hypothesis;Ha: p ≠ 0.68

np = 144 × 0.6548 > 5 and nq = n(1 - p) = 144(1 - 0.68) > 5.

Since np and nq are more than 5, we can use the z-test .

z-formula in this case is;

z = (p^ - p)/√(p(1 - p))

Plugging in the relevant values, we have;

z = (0.6548 - 0.68)/√(0.68(1 - 0.6548))

z = -0.0252/0.4845

z = -0.052

From the z-table attached, we have a p-value of approximately 0.4801

P-value is higher than the significance level of 0.01.

Thus,we will fail to reject the null hypothesis and conclude that there is no sufficient evidence to support the researchers claim.

step-by-step explanation:

(s)he  is right it would be 5.5% because it is smaller than 15.5% or anything smaller than 15.5% in general

i can answer this question sweetie!

step-by-step explanation: