Question 1192384
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Part a)


Hypothesis:
H0: *[tex \large \mu_d = 0]
H1: *[tex \large \mu_d \not=  0]
This is a two-tailed test.



Given Data<table border = "1" cellpadding = "5"><tr><td>before</td><td>after</td></tr><tr><td>29</td><td>39</td></tr><tr><td>22</td><td>26</td></tr><tr><td>25</td><td>25</td></tr><tr><td>29</td><td>35</td></tr><tr><td>26</td><td>33</td></tr><tr><td>24</td><td>36</td></tr><tr><td>31</td><td>32</td></tr></table>


For each row separately, subtract the values in the order A-B to form a column of differences (d)<table border = "1" cellpadding = "5"><tr><td>A</td><td>B</td><td>A-B</td></tr><tr><td>before</td><td>after</td><td>d = difference</td></tr><tr><td>29</td><td>39</td><td>-10</td></tr><tr><td>22</td><td>26</td><td>-4</td></tr><tr><td>25</td><td>25</td><td>0</td></tr><tr><td>29</td><td>35</td><td>-6</td></tr><tr><td>26</td><td>33</td><td>-7</td></tr><tr><td>24</td><td>36</td><td>-12</td></tr><tr><td>31</td><td>32</td><td>-1</td></tr></table>Using a calculator, compute the sample mean and sample standard deviation of the list of difference values in that new column.
You should get *[tex \large \overline{d} \approx -5.714286] for the sample mean and *[tex \large s_d \approx 4.423961] for the sample standard deviation.


The sample size is n = 7 which means the degrees of freedom (df) are
df = n-1
df = 7-1
df = 6
Refer to a T table like this one
<a href = "https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf">https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf</a>
Locate the row that has df = 6
Locate the column that has 0.05 as the two tails
Intersecting the row and column yields the approximate value t = 2.447
<img width="50%" src = "https://i.imgur.com/oM6S2LG.png">


Now let's compute the margin of error.
E = margin of error


*[tex \large E = t*\frac{s_d}{\sqrt{n}}]


*[tex \large E \approx 2.447*\frac{4.423961}{\sqrt{7}}]


*[tex \large E \approx 4.091629]


We can then find the lower bound L
*[tex \large L = \overline{d} - E]


*[tex \large L \approx -5.714286 - 4.091629]


*[tex \large L \approx -9.805915]


*[tex \large L \approx -9.81]


And the upper bound (U) as well
*[tex \large U = \overline{d} + E]


*[tex \large U \approx -5.714286 + 4.091629]


*[tex \large U \approx -1.622657]


*[tex \large U \approx -1.62]


The 95% confidence interval in the format (L, U) is roughly (-9.81, -1.62)
Negative values are valid because it represents the idea that the "after" weights are larger than the "before" ones.


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Part b)


The result of part a) is equivalent to saying we are 95% confident that *[tex \large -9.81 < \mu_d < -1.62] when using the format *[tex \large L < \mu_d < U]
In other words, we are 95% confident that the pameter we're after (*[tex \large \mu_d]) is somewhere between -9.81 and -1.62


Is *[tex \large \mu_d = 0] in this confidence interval? No, it is not. 
So it's unlikely that *[tex \large \mu_d = 0] is the case.
There appears to be a significant difference in the weights.


This means we'd reject the null and side with the alternative hypothesis to conclude that *[tex \large \mu_d \not= 0] 
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