A cognitive psychologist believes that a particular drug improves short-term memory. An experiment is conducted in which 8 randomly selected participants are given the drug and then given a short time to memorize a list of 10 words. Participants are then tested for retention 15 minutes after the memorization period. The number of words correctly recalled by each participant is as follows: 8, 9, 10, 6, 8, 7, 9, 7. Over the past few years, the psychologist has collected a lot of data using this task with similar subjects. Although he has lost the original data, he remembers that the mean was 6 words correctly recalled and that the data were normally distributed.

Required:
a. On the basis of these data, what can we conclude about the effect of the drug on short-term memory? Use α = 0.05.
b. Determine the size of the effect.

We can conclude that there is significant evidence that the drug improves short term memory.

Step-by-step explanation:

Given the sample data : 8, 9, 10, 6, 8, 7, 9, 7.

The sample mean, xbar = Σx / n

Sample standard deviation :

√Σ(x - xbar)²/ n-1)

Xbar = 8

Sample size, n = 8

Standard deviation, s = 1.31

H0 : μ = 6

H0 : μ > 6

The test statistic :

(xbar - μ) ÷ (s/√(n))

(8 - 6) / (1.31 / √8)

2 /0.4631549

Test statistic = 4.318

Decision region :

Reject H0 ; if Pvalue < α

α = 0.05

Pvalue from Tscore calculator :

df = n - 1 ; 8 - 1 =7

Pvalue(4.318, 7) = 0.001744

Pvalue < α ; We reject H0 ; and conclude that there is significant evidence that the drug improves short term memory.

rick would give her the best deal as he will charge $8.88 per hour

answer: 42

use a calculator to find that 845/73 = 11.575

the whole part of the answer is 11. multiply this by 73 to get 11*73 = 803.

subtract that result from 845 to get 845 - 803 = 42