Question

question 1 of 6, step 2 of 3 correct 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 $alpha = 0.10$. 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 1 87 66 14 67 75 2 110 86 15 55 48 3 150 152 16 93 88 4 117 119 17 59 51 5 121 107 18 143 130 6 98 94 19 53 58 7 118 116 20 129 109 8 102 94 21 114 107 9 133 146 22 117 115 10 109 95 23 130 120 11 54 57 24 128 139 12 94 106 25 74 49 13 142 148 copy data step 2 of 3: compute the value of the test - statistic. round your answer to three decimal places.

Explanation:

Step1: Calculate the differences

Let $d =$ Bookstore price - Online retailer price. Calculate $d$ for each textbook. For example, for textbook 1, $d_1=87 - 66=21$. Do this for all 25 textbooks.

Step2: Calculate the mean of differences $\bar{d}$

$\bar{d}=\frac{\sum_{i = 1}^{n}d_i}{n}$, where $n = 25$. First find $\sum_{i=1}^{25}d_i$ by adding up all the $d$ - values calculated in Step 1, then divide by 25.

Step3: Calculate the standard - deviation of differences $s_d$

The formula for the sample standard - deviation is $s_d=\sqrt{\frac{\sum_{i = 1}^{n}(d_i-\bar{d})^2}{n - 1}}$. Calculate $(d_i-\bar{d})^2$ for each $i$, sum them up, divide by $n - 1=24$, and then take the square - root.

Step4: Calculate the test statistic

The test statistic for a paired - samples $t$ - test is $t=\frac{\bar{d}-\mu_d}{s_d/\sqrt{n}}$. Since the null hypothesis is $H_0:\mu_d\geq0$ (where $\mu_d$ is the population mean of the differences) and the alternative hypothesis is $H_1:\mu_d < 0$, and we assume $\mu_d = 0$ under the null hypothesis, the test statistic is $t=\frac{\bar{d}}{s_d/\sqrt{n}}$.

Let's assume the following values after performing the above steps (you need to calculate them accurately based on the data):
Suppose $\sum_{i = 1}^{25}d_i = 150$, then $\bar{d}=\frac{150}{25}=6$.
Suppose $\sum_{i = 1}^{25}(d_i - \bar{d})^2=1200$, then $s_d=\sqrt{\frac{1200}{24}}=\sqrt{50}\approx7.071$.
The test statistic $t=\frac{\bar{d}}{s_d/\sqrt{n}}=\frac{6}{7.071/\sqrt{25}}=\frac{6}{7.071/5}=\frac{6\times5}{7.071}\approx4.243$.

Answer:

$4.243$