Question

The following table contains information on matched sample values whose differences are normally distributed. Use Table 2.

Number	Sample 1	Sample 2
1	16	20
2	12	13
3	20	22
4	20	22
5	17	20
6	14	16
7	16	18
8	18	21

a.	Construct the 99% confidence interval for the mean difference μD. (Negative values should be indicated by a minus sign. Round all intermediate calculations to at least 4 decimal places. Round your answers to 2 decimal places.)

Confidence interval is  to

b.	Specify the competing hypotheses in order to test whether the mean difference differs from zero.
	H0: μD ≥ 0; HA: μD < 0 H0: μD = 0; HA: μD ≠ 0 H0: μD ≤ 0; HA: μD > 0

c.	Using the confidence interval from part a, are you able to reject H0?
	No Yes

Answer 1

Differences (Sample 1 - Sample 2): We calculate sample mean and std deviation from given data Sample Mean, j_ 2d_-19 Sample Variance, s2- Sample std dev, s-V82-V0.839286 0.916125 82.375 Γ(d-d). 5.875 - 0.839286 99% CI for μD using t-dist Sample Mean -d-2.375 Sample Standard deviations0.916125 Sample Size -n- 8 Significance level-α-1-0.99-0.01 Degrees of freedom for t-distribution, d.f.-n -17 Critical Value - ta/2df to.oo5,df 3.499 (from t-table, two-tails, d.f. 7) Margin of Error - E- to/2.df >x E= 3.499 × 0.323899 3.499 0016125 V8 Margin of Error, E1.133323 Limits of 99% confidence interval are given by: Lower limit d- E--2.375 -1.1333233.508323-3.508

Upper limit dE-2.375 1.133323-1.2416771.242 99% confidence interval is: d E--2.375 1.133323 = (-3.508323,-1.241677) 99% CI using t-dist -3.51 < μD < -1.24 Note that exact answer using technology is: -2.375-0.916125381312904/SQRT (8) *T.INV (1-(1-0.99)/2,8-1, -2.3750.916125381312904/SQRT (8)*T.INV(1-(1-0.99)/2,8-1)) -3.5084799621 , -1.2415200379) b) Hypothesis test: Ho : μ = 150 Ha : PDチ150 This is two tailed test. (Null Hypothesis) (Alternative Hypothesis, also called Hı) c Yes ,we reject null because the CI does not include zero. oThank youes