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question 2 of 6, step 2 of 3
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students at a major university believe they can save money buying textbooks online rather than at the local bookstores. in order to test this theory, they randomly sampled 25 textbooks on the shelves of the local bookstores. the students then found the \best\ available price for the same textbooks via online retailers. the prices for the textbooks are listed in the following table. based on the data, is it less expensive for the students to purchase textbooks from the online retailers than from local bookstores? use \\( \alpha = 0.05 \\). let prices at local bookstores represent population 1 and prices at online retailers represent population 2.

textbook prices (dollars)

textbook	bookstore	online retailer	textbook	bookstore	online retailer
2	149	162	15	70	68
3	60	52	16	121	118
4	98	76	17	108	110
5	93	70	18	115	107
6	144	142	19	78	57
7	127	141	20	140	145
8	60	64	21	69	58
9	149	146	22	150	126
10	93	105	23	148	158
11	109	105	24	86	70
12	148	123	25	65	66
13	93	106

step 2 of 3 : compute the value of the test statistic. round your answer to three decimal places.
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Explanation:

Step 1: Calculate the differences \( d_i \)

For each textbook, calculate \( d_i = \text{Bookstore Price} - \text{Online Retailer Price} \).
- Textbook 1: \( 130 - 123 = 7 \)
- Textbook 2: \( 149 - 162 = -13 \)
- Textbook 3: \( 60 - 52 = 8 \)
- Textbook 4: \( 98 - 76 = 22 \)
- Textbook 5: \( 93 - 70 = 23 \)
- Textbook 6: \( 144 - 142 = 2 \)
- Textbook 7: \( 127 - 141 = -14 \)
- Textbook 8: \( 60 - 64 = -4 \)
- Textbook 9: \( 149 - 146 = 3 \)
- Textbook 10: \( 93 - 105 = -12 \)
- Textbook 11: \( 109 - 105 = 4 \)
- Textbook 12: \( 148 - 123 = 25 \)
- Textbook 13: \( 93 - 106 = -13 \)
- Textbook 14: \( 128 - 140 = -12 \)
- Textbook 15: \( 70 - 68 = 2 \)
- Textbook 16: \( 121 - 118 = 3 \)
- Textbook 17: \( 108 - 110 = -2 \)
- Textbook 18: \( 115 - 107 = 8 \)
- Textbook 19: \( 78 - 57 = 21 \)
- Textbook 20: \( 140 - 145 = -5 \)
- Textbook 21: \( 69 - 58 = 11 \)
- Textbook 22: \( 150 - 126 = 24 \)
- Textbook 23: \( 148 - 158 = -10 \)
- Textbook 24: \( 86 - 70 = 16 \)
- Textbook 25: \( 65 - 66 = -1 \)

Step 2: Calculate the mean of the differences \( \bar{d} \)

First, sum all the \( d_i \) values:
\[ \begin{align*} 7 + (-13) + 8 + 22 + 23 + 2 + (-14) + (-4) + 3 + (-12) + 4 + 25 + (-13) + (-12) + 2 + 3 + (-2) + 8 + 21 + (-5) + 11 + 24 + (-10) + 16 + (-1) &= 7 - 13 + 8 + 22 + 23 + 2 - 14 - 4 + 3 - 12 + 4 + 25 - 13 - 12 + 2 + 3 - 2 + 8 + 21 - 5 + 11 + 24 - 10 + 16 - 1 \\ &= (7 + 8 + 22 + 23 + 2 + 3 + 4 + 25 + 2 + 3 + 8 + 21 + 11 + 24 + 16) + (-13 - 14 - 4 - 12 - 13 - 12 - 2 - 5 - 10 - 1) \\ &= 178 + (-76) \\ &= 102 \end{align*} \]
The sample size \( n = 25 \), so the mean \( \bar{d} = \frac{102}{25} = 4.08 \).

Step 3: Calculate the standard deviation of the differences \( s_d \)

First, calculate the squared differences \( (d_i - \bar{d})^2 \) for each \( d_i \):
- For \( d_1 = 7 \): \( (7 - 4.08)^2 = (2.92)^2 = 8.5264 \)
- For \( d_2 = -13 \): \( (-13 - 4.08)^2 = (-17.08)^2 = 291.7264 \)
- For \( d_3 = 8 \): \( (8 - 4.08)^2 = (3.92)^2 = 15.3664 \)
- For \( d_4 = 22 \): \( (22 - 4.08)^2 = (17.92)^2 = 321.1264 \)
- For \( d_5 = 23 \): \( (23 - 4.08)^2 = (18.92)^2 = 357.9664 \)
- For \( d_6 = 2 \): \( (2 - 4.08)^2 = (-2.08)^2 = 4.3264 \)
- For \( d_7 = -14 \): \( (-14 - 4.08)^2 = (-18.08)^2 = 326.8864 \)
- For \( d_8 = -4 \): \( (-4 - 4.08)^2 = (-8.08)^2 = 65.2864 \)
- For \( d_9 = 3 \): \( (3 - 4.08)^2 = (-1.08)^2 = 1.1664 \)
- For \( d_{10} = -12 \): \( (-12 - 4.08)^2 = (-16.08)^2 = 258.5664 \)
- For \( d_{11} = 4 \): \( (4 - 4.08)^2 = (-0.08)^2 = 0.0064 \)
- For \( d_{12} = 25 \): \( (25 - 4.08)^2 = (20.92)^2 = 437.6464 \)
- For \( d_{13} = -13 \): \( (-13 - 4.08)^2 = (-17.08)^2 = 291.7264 \)
- For \( d_{14} = -12 \): \( (-12 - 4.08)^2 = (-16.08)^2 = 258.5664 \)
- For \( d_{15} = 2 \): \( (2 - 4.08)^2 = (-2.08)^2 = 4.3264 \)
- For \( d_{16} = 3 \): \( (3 - 4.08)^2 = (-1.08)^2 = 1.1664 \)
- For \( d_{17} = -2 \): \( (-2 - 4.08)^2 = (-6.08)^2 = 36.9664 \)
- For \( d_{18} = 8 \): \( (8 - 4.08)^2 = (3.92)^2 = 15.3664 \)
- For \( d_{19} = 21 \): \( (21 - 4.08)^2 = (16.92)^2 = 286.2864 \)
- For \( d_{20} = -5 \): \( (-5 - 4.08)^2 = (-9.08)^2 = 82.4464 \)
- For \( d_{21} = 11 \): \( (11 - 4.08)^2 = (6.92)^2 = 47.8864 \)
- For \( d_{22} = 24 \): \( (24 - 4.08)^2 = (19.92)^2 = 396.8064 \)
- For \( d_{23} = -10 \): \( (-10 - 4.08)^2 = (-14.08)^2 = 198.2464 \)
- For \( d_{24} = 16 \): \( (16 - 4.08)^2 = (11.92)^2 = 142.0864 \)
- For \( d_{25} = -1 \): \( (-1 - 4.08)^2 = (-5.08)^2 = 25.8064 \)

Now, sum all these squared differences:
\[ \begin{align*} &8.5264 + 291.7264 + 15.3664 + 321.1264 + 357.9664 + 4.3264 + 326.8864 + 65.2864 + 1.1664 + 258.5664 + 0.0064 + 437.6464 + 291.7264 + 258.5664 + 4.3264 + 1.1664 + 36.9664 + 15.3664 + 286.2864 + 82.4464 + 47.8864 + 396.8064 + 198.2464 + 142.0864 + 25.8064 \\ &= 4333.36 \end{align*} \]
The sample standard deviation \( s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n - 1}} = \sqrt{\frac{4333.36}{24}} \approx \sqrt{180.5567} \approx 13.437 \).

Step 4: Calculate the test statistic \( t \)

The formula for the paired - t test statistic is \( t=\frac{\bar{d}-\mu_d}{s_d/\sqrt{n}} \). Here, we are testing if the online prices are less expensive, so the null hypothesis is \( H_0:\mu_d = 0 \) and the alternative hypothesis is \( H_1:\mu_d>0 \) (since \( \mu_d=\mu_1 - \mu_2 \), and we want to test if \( \mu_2<\mu_1 \) or \( \mu_1-\mu_2 > 0 \)).
Substitute \( \bar{d} = 4.08 \), \( s_d\approx13.437 \), \( n = 25 \) and \( \mu_d = 0 \) into the formula:
\[ t=\frac{4.08-0}{13.437/\sqrt{25}}=\frac{4.08}{13.437/5}=\frac{4.08\times5}{13.437}=\frac{20.4}{13.437}\approx1.518 \]

Answer:

\( 1.518 \)