Question

The accompanying table contains the service ratings of 14 different Internet and TV providers. Complete parts (a) through(d) below

At the 0.01 level of significance, is there evidence of a difference in the mean service rating between TV and Internetservices?Let mu 1be the mean TV service rating and let mu 2 be the mean Internet service rating. State the null and alternative hypotheses. Choose the correct answer below.

b.Determine the T Test and critical values

c. Determine the P value

d Construct and interpret a 90% confidence interval estimate of the difference in the mean service rating between TV and Internet services aswell as a 95% confidence interval estimate

Provider TV Internet

1 65 69

2 57 59

3 71 75

4 67 71

5 67 69

6 63 66

7 55 56

8 64 66

9 76 77

10 74 77

11 75 77

12 68 73

13 64 67

14 61 65

Answer #1

**A)**

H0= Both means are equal

H1=Means are not equal

Using a 0.01 level of significance

Two-Sample T-Test and CI: TV, Internet

Two-sample T for TV vs Internet

N Mean StDev SE Mean

TV 14 66.21 6.33 1.7

Internet 14 69.07 6.53 1.7

Difference = μ (TV) - μ (Internet)

Estimate for difference: -2.86

99% CI for difference: (-9.63, 3.92)

T-Test of difference = 0 (vs ≠):

DF = 25

**b)**

T-Value = -1.18

**C)**

P-Value = 0.251

Here P value is greater than 0.05 so we accept H0 so there is no difference between this two Tv and internet both means are same.

**d)**

90% confidence interval estimate of the difference in the mean service rating between TV and Internet services

=90% CI for difference: (-7.01, 1.29)

95% confidence interval estimate of the difference in the mean service rating between TV and Internet services

=95% CI for difference: (-7.86, 2.15)

Refer to the accompanying data set and construct a
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females; then do the same for adult males. Compare the results.
Click the icon to view the pulse rates for adult females and
adult males.
Construct a 95% confidence interval of the mean pulse rate for
adult females.nothing
bpm<μ<bpm
bpm
(Round to one decimal place as needed.)
Construct a
95%
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81
82
72
94
52
57
59
66
51
54
62
80
52
77
74
85
51
89
62
57
73
35
61
64
62
87
80
74
80
79
63
62
63
67
97
76
43
60
85
65
72
86
66
85
73
69
72 ...

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Final Round
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70
66
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67
77
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71
65
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73
72
Soren Hansen
73
67
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77
69
Bubba Watson
77
72
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67
68
Jeff Klauk
66
65
Kenny Perry
76
68
Aron Price
68
66
Charles Howell
72
69...

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Let d=(this year's rating)−(last year's rating). Assume that the
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year. The following table shows the legislator's performance from
the same ten randomly selected voters for last year and this year.
Use this data to find the 90% confidence interval for the true
difference between the population means. Assume that the
populations of voters' performance ratings are normally distributed
for both this year and last year. Rating (last year)...

Notes:
Please type your work, or 10% credits will be deducted
Show your work for full credits
Follow the six steps for hypothesis testing
If Excel is used, please copy and paste Excel Output as part of
your answer
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