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.

c

step-by-step explanation:

see attached

step-by-step explanation:

given the function as;

write the equation replacing f(x) with y

form   and fill a table containing values of y with corresponding values of x

you can take a range for   values of x as -3≤x≤3

solving for x=-3

for x=-2

repeat the same procedure to get the values of y for the remaining ,x=-1 to x=3

plot the values of y against values   of x and obtain the following graph