“I can’t get through my day without coffee” is a common statement from many students. Assumed benefits include keeping students awake during lectures and making them more alert for exams and tests. Students in a statistics class designed an experiment to measure memory retention with and without drinking a cup of coffee one hour before a test. This experiment took place on two different days in the same week (Monday and Wednesday). Ten students were used. Each student received no coffee or one cup of coffee, one hour before the test on a particular day. The test consisted of a series of words flashed on a screen, after which the student had to write down as many of the words as possible. On the other day, each student received a different amount of coffee (none or one cup). (a) One of the researchers suggested that all the subjects in the experiment drink no coffee before Monday’s test and one cup of coffee before Wednesday’s test. Explain to the researcher why this is a bad idea and suggest a better method of deciding when each subject receives the two treatments.

(b) The data from the experiment are provided in the table below. Set up and carry out an appropriate test to determine whether there is convincing evidence that drinking coffee improves memory.

Step by step solution

01

Part (a) Step 1: Given Information

02

Part (a) Step 2: Explanation

Students are likely to perform better on Wednesday's test than on Monday's test due to experience and thus it would be better to randomly assign each student to drink coffee before the test either on Monday or on Wednesday.

03

Part (b) Step 1: Given Information

04

Part (b) Step 2: Explanation 

Determine the difference in the score for each student:

05

Part (b) Step 3: Calculate the mean and standard deviation

The mean is the sum of all values divided by the number of values:

$\overline{x}=\frac{-1-1-1+0-1-2-2+0+0-2}{10}$

$=-1$

$n$is the number of values in the data set.

The variance is the sum of squared deviations from the mean divided by $n-1$:

localid="1650365466814" $s^2=\frac{(-1-(-1){)}^2+\dots .+(-2-(-1){)}^2}{10-1}$

$\approx 0.6$

The standard deviation is the square root of the variance:

localid="1650365483106" $s=\sqrt{0.6}$

$\approx 0.8165$

06

Part (b) Step 4: Compute the test statistics

$H_0:\mu =0$

$H_a:\mu <0$

Determine the value of the test statistic:

localid="1650365512177" $t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}$

$=\frac{-1-0}{0.8165/\sqrt{10}}$

$=-3.873$

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Table B containing the t-value in the row localid="1650365551205" $n-1=10-1$

$=9$ :

$0.001<P<0.0025$

If the P-value is smaller than the significance level, then the null hypothesis is rejected.

$P<0.05=5\%\Rightarrow \text{Reject}{H}_0$

There is sufficient evidence to support the claim that drinking coffee improves memory.

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