Show that in Example 4.4.7, page 153, the difference between the two candidates is statistically significant even at a confidence level of 99.7%.

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Show that in Example 4.4.7, page 153, the difference between the two candidates is statistically significant even at a confidence level of 99.7%.

Now to the numbers! Let's begin with Candidate A. We have the following
numbers: sample size n =
1350, favorable voters k
648. Therefore p
648
0.48 or 48%, and o 2
1350 x 0.48 х (1 — 0.48) ~ 18.3565, so the
1350
18.3565
standard error is st.err. -
a 0.0136 or 1.36%.
1350
CHAPTER 4. STATISTICS
153
This tells us that the percentage of support for Candidate A the day of
the poll was between 48 – 2 × 1.36 and 48+2×1.36, that is between 45.28%
and 50.72%.
Similarly, for Candidate B we have n =
1350, favorable voters k
702.
702
Therefore p =
— 0.52 or 52%, and o ~ y/1350 x 0.52 х (1 — 0.52) ~
1350
18.3565
18.3565, so the standard error is st.err. 2
2 0.0136 or 1.36%.
1350
This tells us that the percentage of support for Candidate B the day of
the poll was between 52 – 2 × 1.36 and 52+ 2×1.36, that is between 49.28%
and 54.72%.
When we plot these two intervals one on top the other we clearly see that
they overlap.
A
E**
*****I
45.28
48
50.72
В
E******:
******I
49.28
52
54.72
Overlap
E********I
49.28
50.72
Figure 4.18: Overlapping Confidence Intervals for Two Candidates
That means that with a 95% confidence level, we cannot say that one
candidate beats the other. We call this situation a statistical tie, or we say
Transcribed Image Text:Now to the numbers! Let's begin with Candidate A. We have the following numbers: sample size n = 1350, favorable voters k 648. Therefore p 648 0.48 or 48%, and o 2 1350 x 0.48 х (1 — 0.48) ~ 18.3565, so the 1350 18.3565 standard error is st.err. - a 0.0136 or 1.36%. 1350 CHAPTER 4. STATISTICS 153 This tells us that the percentage of support for Candidate A the day of the poll was between 48 – 2 × 1.36 and 48+2×1.36, that is between 45.28% and 50.72%. Similarly, for Candidate B we have n = 1350, favorable voters k 702. 702 Therefore p = — 0.52 or 52%, and o ~ y/1350 x 0.52 х (1 — 0.52) ~ 1350 18.3565 18.3565, so the standard error is st.err. 2 2 0.0136 or 1.36%. 1350 This tells us that the percentage of support for Candidate B the day of the poll was between 52 – 2 × 1.36 and 52+ 2×1.36, that is between 49.28% and 54.72%. When we plot these two intervals one on top the other we clearly see that they overlap. A E** *****I 45.28 48 50.72 В E******: ******I 49.28 52 54.72 Overlap E********I 49.28 50.72 Figure 4.18: Overlapping Confidence Intervals for Two Candidates That means that with a 95% confidence level, we cannot say that one candidate beats the other. We call this situation a statistical tie, or we say
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