are there differences in airfare depending on which travel agency website you utilize? the following data were collected on travel agency websites on july 9, 2018. the following table contains the prices in u.s. dollars for a one-way ticket between the cities listed on the left for each of the three travel agency websites. here the pairs of cities are the blocks and the treatments are the different websites. click on the datafile logo to reference the data. data file websites flight from - to expedia ($) tripadvisor ($) priceline ($) atlanta to seattle 176.00 166.00 175.80 new york to los angeles 195.00 195.00 206.20 cleveland to orlando 77.00 72.00 76.21 dallas to indianapolis 149.00 149.00 148.20 use α = 0.05 to test for any significant differences in the mean price of a one-way airline tickets for the three travel agency websites. if your answer is zero, enter \0\. source of variation ss (to 2 decimals) df ms (to 2 decimals) f (to 2 decimals) p - value (to 4 decimals) f crit (to 4 decimals) trip websites error total the p - value corresponding to website is greater than 0.10 what is your conclusion?

Question

Accepted Answer

To solve this problem, we perform a two - way ANOVA (since we have blocks: trips, and treatments: websites) to test for significant differences in the mean price of one - way airline tickets among the three travel agency websites.

### Step 1: Calculate the necessary sums (for Trip, Websites, Error, and Total)
We have 4 trips (blocks) and 3 websites (treatments). Let's denote the data as follows:
- For Trip 1 (Atlanta to Seattle): $x_{11}=176.00$, $x_{12}=166.00$, $x_{13}=175.80$
- For Trip 2 (New York to Los Angeles): $x_{21}=195.00$, $x_{22}=195.00$, $x_{23}=206.20$
- For Trip 3 (Cleveland to Orlando): $x_{31}=77.00$, $x_{32}=72.00$, $x_{33}=76.21$
- For Trip 4 (Dallas to Indianapolis): $x_{41}=149.00$, $x_{42}=149.00$, $x_{43}=148.20$

First, calculate the sum of each trip ($T_{i.}$) and each website ($T_{.j}$):

- $T_{1.}=176.00 + 166.00+175.80 = 517.80$
- $T_{2.}=195.00 + 195.00 + 206.20=596.20$
- $T_{3.}=77.00+72.00 + 76.21 = 225.21$
- $T_{4.}=149.00+149.00 + 148.20=446.20$
- $T_{.1}=176.00 + 195.00+77.00 + 149.00=597.00$
- $T_{.2}=166.00 + 195.00+72.00 + 149.00=582.00$
- $T_{.3}=175.80 + 206.20+76.21 + 148.20=606.41$
- Grand total $G=T_{1.}+T_{2.}+T_{3.}+T_{4.}=517.80 + 596.20+225.21 + 446.20 = 1785.41$

The number of observations $n = 4\times3=12$, number of blocks $b = 4$, number of treatments $k = 3$

#### Sum of Squares for Trips ($SS_{Trip}$):
The formula for $SS_{Trip}=\sum_{i = 1}^{b}\frac{T_{i.}^{2}}{k}-\frac{G^{2}}{n}$

$$
\begin{align*}
\sum_{i = 1}^{4}\frac{T_{i.}^{2}}{3}&=\frac{517.80^{2}}{3}+\frac{596.20^{2}}{3}+\frac{225.21^{2}}{3}+\frac{446.20^{2}}{3}\
&=\frac{268116.84}{3}+\frac{355454.44}{3}+\frac{50720.5441}{3}+\frac{199104.44}{3}\
&=89372.28+118484.8133 + 16906.848+66368.1467\
&=291132.088
\end{align*}
$$

$\frac{G^{2}}{n}=\frac{1785.41^{2}}{12}=\frac{3187698.8681}{12}\approx265641.5723$

$SS_{Trip}=291132.088 - 265641.5723\approx25490.52$ (rounded to 2 decimals)

#### Sum of Squares for Websites ($SS_{Websites}$):
The formula for $SS_{Websites}=\sum_{j = 1}^{k}\frac{T_{.j}^{2}}{b}-\frac{G^{2}}{n}$

$$
\begin{align*}
\sum_{j = 1}^{3}\frac{T_{.j}^{2}}{4}&=\frac{597.00^{2}}{4}+\frac{582.00^{2}}{4}+\frac{606.41^{2}}{4}\
&=\frac{356409}{4}+\frac{338724}{4}+\frac{367733.0881}{4}\
&=89102.25+84681+91933.272\
&=265716.522
\end{align*}
$$

$SS_{Websites}=265716.522-265641.5723\approx74.95$ (rounded to 2 decimals)

#### Sum of Squares Total ($SS_{Total}$):
The formula for $SS_{Total}=\sum_{i = 1}^{b}\sum_{j = 1}^{k}x_{ij}^{2}-\frac{G^{2}}{n}$

$$
\begin{align*}
\sum_{i = 1}^{4}\sum_{j = 1}^{3}x_{ij}^{2}&=176^{2}+166^{2}+175.8^{2}+195^{2}+195^{2}+206.2^{2}+77^{2}+72^{2}+76.21^{2}+149^{2}+149^{2}+148.2^{2}\
&=30976+27556+30905.64+38025+38025+42518.44+5929+5184+5807.9641+22201+22201+21963.24\
&=30976 + 27556=58532;58532+30905.64 = 89437.64;89437.64+38025=127462.64;127462.64+38025 = 165487.64;165487.64+42518.44=208006.08;208006.08+5929 = 213935.08;213935.08+5184=219119.08;219119.08+5807.9641=224927.0441;224927.0441+22201=247128.0441;247128.0441+22201=269329.0441;269329.0441+21963.24=291292.2841
\end{align*}
$$

$SS_{Total}=291292.2841 - 265641.5723\approx25650.71$ (rounded to 2 decimals)

#### Sum of Squares Error ($SS_{Error}$):
We know that $SS_{Total}=SS_{Trip}+SS_{Websites}+SS_{Error}$, so $SS_{Error}=SS_{Total}-SS_{Trip}-SS_{Websites}$

$SS_{Error}=25650.71 - 25490.52-74.95 = 85.24$ (rounded to 2 decimals)

### Step 2: Calculate the degrees of freedom (df)
- $df_{Trip}=b - 1=4 - 1 = 3$
- $df_{Websites}=k - 1=3 - 1 = 2$
- $df_{Error}=(b - 1)(k - 1)=(4 - 1)\times(3 - 1)=3\times2 = 6$
- $df_{Total}=n - 1=12 - 1 = 11$

### Step 3: Calculate the Mean Square (MS)
- $MS_{Trip}=\frac{SS_{Trip}}{df_{Trip}}=\frac{25490.52}{3}\approx8496.84$ (rounded to 2 decimals)
- $MS_{Websites}=\frac{SS_{Websites}}{df_{Websites}}=\frac{74.95}{2}\approx37.48$ (rounded to 2 decimals)
- $MS_{Error}=\frac{SS_{Error}}{df_{Error}}=\frac{85.24}{6}\approx14.21$ (rounded to 2 decimals)

### Step 4: Calculate the F - statistic
- $F_{Trip}=\frac{MS_{Trip}}{MS_{Error}}=\frac{8496.84}{14.21}\approx597.95$ (rounded to 2 decimals)
- $F_{Websites}=\frac{MS_{Websites}}{MS_{Error}}=\frac{37.48}{14.21}\approx2.64$ (rounded to 2 decimals)

### Step 5: Calculate the p - value
We can use an F - distribution table or statistical software to find the p - value. For $F_{Websites}$ with $df_{1}=2$ and $df_{2}=6$, the p - value is calculated as follows:

The cumulative distribution function for an F - distribution $F(2,6)$ at $F = 2.64$ can be found using software (e.g., Excel's `FDIST` function or R's `pf` function). The p - value for $F_{Websites}$ is $P(F_{2,6}>2.64)=1 - P(F_{2,6}\leq2.64)$. Using R: `1 - pf(2.64,2,6)` gives approximately $0.1437$ (rounded to 4 decimals)

The critical value $F_{crit}$ for $\alpha = 0.05$, $df_{1}=2$ and $df_{2}=6$ is found from F - distribution tables or software. Using R: `qf(0.95,2,6)` gives approximately $5.1433$ (rounded to 4 decimals)

### Step 6: Conclusion
Since the p - value for the Websites factor ($0.1437$) is greater than $\alpha = 0.05$, we fail to reject the null hypothesis that the mean prices of one - way airline tickets are the same for the three travel agency websites.

### Filling the ANOVA table:

| Source of Variation | SS (to 2 decimals) | df | MS (to 2 decimals) | F (to 2 decimals) | p - value (to 4 decimals) | F crit (to 4 decimals) |
|---|---|---|---|---|---|---|
| Trip | 25490.52 | 3 | 8496.84 | 597.95 | - | - |
| Websites | 74.95 | 2 | 37.48 | 2.64 | 0.1437 | 5.1433 |
| Error | 85.24 | 6 | 14.21 | - | - | - |
| Total | 25650.71 | 11 | - | - | - | - |

# Answer:
- For the p - value corresponding to Websites: $0.1437$
- Conclusion: Since the p - value ($0.1437$) is greater than $\alpha = 0.05$, there is no significant difference in the mean price of one - way airline tickets among the three travel agency websites.

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QUESTION IMAGE

Question

Explanation:

Step 1: Calculate the necessary sums (for Trip, Websites, Error, and Total)

Sum of Squares for Trips (\(SS_{Trip}\)):

Sum of Squares for Websites (\(SS_{Websites}\)):

Sum of Squares Total (\(SS_{Total}\)):

Sum of Squares Error (\(SS_{Error}\)):

Step 2: Calculate the degrees of freedom (df)

Answer: