The president of a university claimed that the entering class this year appeared to be larger than the entering class from previous years but their mean SAT score is lower than previous years. He took a sample of 20 of this year's entering students and found that their mean SAT score is 1,501 with a standard deviation of 53. The university's record indicates that the mean SAT score for entering students from previous years is 1,520. He wants to find out if his claim is supported by the evidence at a 5% level of significance. Referring to Scenario 9-9, what is the critical value

Critical value = -1.729

Step by Step explanation:

Given:

n = 20

X' = 1501

Standard deviation = 53

Mean, u = 1520

Level of significance, a = 0.05

The null and alternative hypotheses:

H0 : u = 1520

H1 : u < 1520

This is a lower tailed test.

Degrees of freedom, df = 20 - 1 = 19

For critical value:

From t table df = 19, one tailed

Critical value = -1.729

Decision: Reject null hypothesis H0, if test statistic Z, is less than critical value.

Test statistic Z =

Z = -1.603

For p-value:

From excel,

P(t< -1.603) = t.dist( -1.603, 19, 1)

= 0.06269

≈ 0.0627

P value = 0.0627

Since test statistic Z, -1.603, is greater than critical value, -1.729, we fail to reject the null hypothesis H0.

There is not enough statistical evidence to conclude that mean is less than 1520.

15%

step-by-step explanation:

use the table and count the number of guesses that has two correct answers.

the following are two correct answers

10010, 01100, 11000

a total of 3 numbers having 2 correct guesses out of 20 questions.

probability of getting 2 correct guesses = the number of questions getting 2 correct answers / total number of questions.

= 3/20

to convert it into percent multiply by 100, we get

= 3/20 * 100

= 15%

the answer is 15%

you.

1) is right

step-by-step explanation: