Question

Assume that the differences are normally distributed. Complete parts ​(a) through ​(d) below.

Observation	1	2	3	4	5	6	7	8	Open in StatCrunch + Copy to Clipboard + Open in Excel +
Upper X Subscript iXi	46.746.7	47.747.7	45.645.6	50.250.2	48.448.4	50.850.8	47.847.8	48.648.6
Upper Y Subscript iYi	50.150.1	48.448.4	47.347.3	54.554.5	47.947.9	50.950.9	49.649.6	50.350.3

​(a) Determine

d Subscript i Baseline equals Upper X Subscript i Baseline minus Upper Y Subscript idi=Xi−Yi

for each pair of data.

Observation	1	2	3	4	5	6	7	8
di	negative 3.4 −3.4	negative 0.7 −0.7	negative 1.7 −1.7	negative 4.3 −4.3	0.5 0.5	negative 0.1 −0.1	negative 1.8 −1.8	negative 1.7 −1.7

​(Type integers or​ decimals.)

​(b) Compute

d overbard

and

s Subscript dsd.

d overbardequals=negative 1.650 −1.650

​(Round to three decimal places as​ needed.)

s Subscript dsdequals=1.605 1.605

​(Round to three decimal places as​ needed.)​(c) Test if

mu Subscript dμdless than<0

at the

alphaαequals=0.05

level of significance.

What are the correct null and alternative​ hypotheses?

A.

Upper H 0H0​:

mu Subscript dμdless than<0

Upper H 1H1​:

mu Subscript dμdequals=0

B.

Upper H 0H0​:

mu Subscript dμdgreater than>0

Upper H 1H1​:

mu Subscript dμdless than<0

C.

Upper H 0H0​:

mu Subscript dμdless than<0

Upper H 1H1​:

mu Subscript dμdgreater than>0

D.

Upper H 0H0​:

mu Subscript dμdequals=0

Upper H 1H1​:

mu Subscript dμdless than<0

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_d> 0

Alternative hypothesis: u_d < 0

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

s = 1.60535

SE = s / sqrt(n)

S.E = 0.5676

DF = n - 1 = 8 -1

D.F = 7

t = [ (x₁ - x₂) - D ] / SE

t = - 2.907

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a one-tailed test, the P-value is the probability that a t statistic having 7 degrees of freedom is less than - 2.907.

Thus, the P-value = 0.011

Interpret results. Since the P-value (0.011) is less than the significance level (0.05), we have to reject the null hypothesis.

Reject H0.