Question

Consider the following sample data drawn independently from normally distributed populations with equal population variances. Use Table 2. Sample 1 Sample 2 11.0 9.3 10.8 11.9 7.3 12.5 12.5 11.4 10.6 9.7 9.8 10.0 7.2 12.6 10.5 12.7 Click here for the Excel Data File a. Construct the relevant hypotheses to test if the mean of the second population is greater than the mean of the first population. H0: μ1 − μ2 = 0; HA: μ1 − μ2 ≠ 0 H0: μ1 − μ2 ≥ 0; HA: μ1 − μ2 < 0 H0: μ1 − μ2 ≤ 0; HA: μ1 − μ2 > 0 b-1. Calculate the value of the test statistic. (Negative values should be indicated by a minus sign. Round all intermediate calculations to at least 2 decimal places and final answer to 2 decimal places.) Test statistic b-2. Calculate the critical value at the 1% level of significance. (Negative value should be indicated by a minus sign. Round your answer to 3 decimal places.) Critical value b-3. Using the critical value approach, can we reject the null hypothesis at the 1% significance level? Yes, since the value of the test statistic is less than the critical value of -2.624. Yes, since the value of the test statistic is less than the critical value of -2.977. No, since the value of the test statistic is not less than the critical value of -2.977. No, since the value of the test statistic is not less than the critical value of -2.624. c. Using the critical value approach, can we reject the null hypothesis at the 5% level? No, since the value of the test statistic is not less than the critical value of -1.761. Yes, since the value of the test statistic is less than the critical value of -1.345. Yes, since the value of the test statistic is less than the critical value of -1.761. No, since the value of the test statistic is not less than the critical value of -1.345.

Answer #1

The statistical software output for this problem is:

Hence,

a) Hypotheses: H0: μ1 − μ2 ≥ 0; HA: μ1 − μ2 < 0

b - 1) Test statistic = -1.14

b - 2) Critical value = -**2.624**

b - 3) No, since the value of the test statistic is not less than the critical value of -2.624.

c) No, since the value of the test statistic is not less than the critical value of -1.761.

Consider the following sample data drawn independently from
normally distributed populations with equal population
variances.
Sample 1
Sample 2
11.2
11.4
11.5
12.1
7.7
12.7
10.7
10.2
10.2
10.2
9.1
9.9
9.3
10.9
11.6
12.7
a. Construct the relevant hypotheses to test if
the mean of the second population is greater than the mean of the
first population.
a) H0: μ1 −
μ2 = 0; HA:
μ1 − μ2 ≠ 0
b) H0: μ1 −
μ2 ≥ 0; HA:
μ1...

Consider the following sample data drawn independently from
normally distributed populations with unknown but equal population
variances. (You may find it useful to reference the
appropriate table: z table or t
table)
Sample 1
Sample 2
12.1
8.9
9.5
10.9
7.3
11.2
10.2
10.6
8.9
9.8
9.8
9.8
7.2
11.2
10.2
12.1
Click here for the Excel Data File
a. Construct the relevant hypotheses to test if
the mean of the second population is greater than the mean of the...

Consider the following competing hypotheses and accompanying
sample data drawn independently from normally distributed
populations. (Note: the automated question following this one will
ask you confidence interval questions for this same data, so jot
down your work.)
H0: μ1 −
μ2 = 0
HA: μ1 −
μ2 ≠ 0
x−1x−1 = 74
x−2x−2 = 65
σ1 = 1.57
σ2 = 14.10
n1 = 19
n2 = 19
a-1.
Calculate the value of the test statistic. (Negative
values should be indicated...

Consider the following data drawn independently from normally
distributed populations:
x1 = 34.4
x2 = 26.4
σ12 = 89.5
σ22 = 95.8
n1 = 21
n2 = 23
a. Construct the 90% confidence interval for the
difference between the population means.
(Negative values should be indicated by a
minus sign. Round all intermediate calculations to at least 4
decimal places and final answers to 2 decimal
places.)
Confidence interval is__________ to__________.
b. Specify the competing hypotheses in order to
determine...

Consider the following competing hypotheses and accompanying
sample data drawn independently from normally distributed
populations. (You may find it useful to reference the
appropriate table: z table or t
table)
H0: μ1 −
μ2 = 0
HA: μ1 −
μ2 ≠ 0
x−1x−1 = 75
x−2x−2 = 79
σ1 = 11.10
σ2 = 1.67
n1 = 20
n2 = 20
a-1. Calculate the value of the test statistic.
(Negative values should be indicated by a minus sign. Round
all intermediate...

Consider the following competing hypotheses and accompanying
sample data drawn independently from normally distributed
populations. (You may find it useful to reference the
appropriate table: z table or t
table)
H0: μ1 −
μ2 = 0
HA: μ1 −
μ2 ≠ 0
x−1x−1 = 57
x−2x−2 = 63
σ1 = 11.5
σ2 = 15.2
n1 = 20
n2 = 20
a-1. Calculate the value of the test statistic.
(Negative values should be indicated by a minus sign. Round
all intermediate...

Consider the following competing hypotheses and accompanying
sample data drawn independently from normally distributed
populations. (You may find it useful to reference the
appropriate table: z table or t
table)
H0: μ1 −
μ2 = 0
HA: μ1 −
μ2 ≠ 0
x−1x−1 = 68
x−2x−2 = 80
σ1 = 12.30
σ2 = 1.68
n1 = 15
n2 = 15
a-1. Calculate the value of the test statistic.
(Negative values should be indicated by a minus sign. Round
all intermediate...

Consider the following data drawn independently from normally
distributed populations: (You may find it useful to
reference the appropriate table: z table
or t table)
x−1x−1 = 28.5
x−2x−2 = 29.8
σ12 = 96.9
σ22 = 87.0
n1 = 29
n2 = 25
a. Construct the 99% confidence interval for the
difference between the population means.
(Negative values should be indicated by a
minus sign. Round all intermediate calculations to at least 4
decimal places and final answers to 2...

Consider the following data drawn independently from normally
distributed populations: (You may find it useful to
reference the appropriate table: z table
or t table)
x−1x−1 = 29.8
x−2x−2 = 32.4
σ12 = 95.3
σ22 = 91.6
n1 = 34
n2 = 29
a. Construct the 99% confidence interval for the
difference between the population means.
(Negative values should be indicated by a
minus sign. Round all intermediate calculations to at least 4
decimal places and final answers to 2...

Consider the following data drawn independently from normally
distributed populations: (You may find it useful to
reference the appropriate table: z table
or t table)
x−1x−1 = 34.4
x−2x−2 = 26.4
σ12 = 89.5
σ22 = 95.8
n1 = 21
n2 = 23
a. Construct the 90% confidence interval for the
difference between the population means.
(Negative values should be indicated by a
minus sign. Round all intermediate calculations to at least 4
decimal places and final answers to 2...

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