A recent study showed that 53% of college applications were submitted online. Assume this result is based on a simple random sample of 1000 college applications, with 530 submitted online. Use a .01 significance level to test the claim that among all college applications, the percentage submitted online is equal to 50%. You may choose to sketch the curve on your scratch paper for visual support.

Step-by-step explanation:

We would set up the hypothesis test. This is a two tailed test

a) For the null hypothesis,

P = 0.5

For the alternative hypothesis,

P ≠ 0.5

Considering the population proportion, probability of success, p = 0.5

q = probability of failure = 1 - p

q = 1 - 0.5 = 0.5

Considering the sample,

Sample proportion, p = x/n

Where

x = number of success = 530

n = number of samples = 1000

p = 530/1000 = 0.53

We would determine the test statistic which is the z score

z = (p - P)/√pq/n

z = (0.53 - 0.50)/√(0.5 × 0.5)/1000 = 1.9

Recall, population proportion, P = 0.5

The difference between sample proportion and population proportion(P - p) is 0.53 - 0.5 = 0.03

Since the curve is symmetrical and it is a two tailed test, the p for the left tail is 0.5 - 0.03 = 0.47

the p for the right tail is 0.5 + 0.03 = 0.53

These proportions are lower and higher than the null proportion. Thus, they are evidence in favour of the alternative hypothesis. We will look at the area in both tails. Since it is showing in one tail only, we would double the area

From the normal distribution table, the area above the z score in the right tail is 1 - 0.9713 = 0.0287

We would double this area to include the area in the left tail of z = - 1.9. Thus

p = 0.0287 × 2 = 0.0574

d) Since alpha, 0.01 < than the p value, 0.0574, then we would fail to reject the null hypothesis

8(-10x+3.5y-7)

8×-10x + 8×3.5y + 8×-7

-80x + 28y - 56

therefore

-80x+28y-56 is the correct answer

2

step-by-step explanation: