N=400 people under the age of 25 was asked whether they check social media sites right after they wake up. 42% check social media sites in the morning and 58% said that they don’t. Using a 5% significance level, test to see if there is evidence that less than 50% of people under the age of 25 check social media sites right after they wake up.

Step-by-step explanation:

We would set up the hypothesis test.

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,

P = 42/100 = 0.42

We would determine the test statistic which is the z score

z = (P - p)/√pq/n

n = 400

z = (0.42 - 0.5)/√(0.5 × 0.5)/400 = - 3.2

Recall, population proportion, p = 0.5

We want the area to the left of 0.5 since the alternative hypothesis is lesser than 0.5. Therefore, from the normal distribution table, the probability of getting a proportion < 0.5 is 0.00069

So p value = 0.00069

Since alpha, 0.05 > than the p value, 0.00069, then we would reject the null hypothesis.

Therefore, there is significant evidence to conclude that that less than 50% of people under the age of 25 check social media sites right after they wake up.

the expected value of the random variable : 0, 1 and 2

step-by-step explanation:

the answer is 9.8975

explanation:

when you multiply the "39.59" with the 25% it reduces it and shows its final discounted price. hope this !

step-by-step explanation: