In 1908, W. S. Gossett performed a sleep study on 10 subjects, comparing the number of hours of sleep obtained by each subject with and without (in that order) the application of the drug hyoscyamine. The paired differences d were as follows: 1.0, 0.8, 1.1, 0.1, - 0.1, 4.4, 1.5, 1.6, 4.6, 3.4 At the 1% significance level, is the drug effective in increasing sleep

Since the calculated value of t= 2.8782 does not fall in the critical region so we accept H0 and may conclude that  the drug is not effective in increasing sleep.

Step-by-step explanation:

d              d²

1.0,          1

0.8,         0.64

1.1,           1.21

0.1,          0.01

- 0.1,       0.01

4.4,        19.36

1.5,          2.25

1.6,          2.56

4.6,         21.16

3.4          11.56

∑18.4      ∑59.76

1: We state our null hypothesis  as

H0 : μd= 0     against the claim Ha: μd ≠ 0  

2: The significance level is set at ∝  = 0.01

3: The test statistic under H0 is

        t= d`/ sd /√n

4:The critical region is t ≥ t ( 0.005) 9 = 3.250

5:Computation:

d`= ∑d/n= 18.4/10= 1.84

Sd² = ∑(di- d`)²/n-1 = 1/n-1 [∑di²- (∑di)² /n]

       = 1/9 [59.76 - (18.4)²/10]

        =(59.76 - 33.856)/9

          = 25.904/9

          = 2.8782

6: Conclusion:

Since the calculated value of t= 2.8782 does not fall in the critical region so we accept H0 and may conclude that  the drug is not effective in increasing sleep.

This is because we have taken H0 as the mean of the difference is zero.

This can only be zero when the drug is not effective.

0.16 or 16%

step-by-step explanation:

lets take a coin for example.

if you toss it twice, it has a 1/4 chance of landing on heads both times because 1/2 * 1/2 is 1/4.

with the cards, the chance of picking dog is 0.4. having two people pick one is like picking a card twice in a row. so 0.4 * 0.4 is 0.16, or 16%