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Math  /  Data & Statistics

QuestionResearchers want to test a new anti-unxicty medication. They split participants into three conditions (0mg,50mg(0 \mathrm{mg}, 50 \mathrm{mg}, and 100 mg)), then ask them to rate their anciety leyel on a scale of 1-10. Compute the value of the tes suatistic. A) F=96.33F=96.33 B) F=86.33F=86.33 C) F=77.33\mathrm{F}=77.33 D) F=67.33\mathrm{F}=67.33 \begin{tabular}{|l|l|l|} \hline Omg & 50 mg & 100 mg \\ \hline 9 & 7 & 4 \\ \hline 8 & 6 & 3 \\ \hline 7 & 6 & 2 \\ \hline 8 & 7 & 3 \\ \hline 8 & 8 & 4 \\ \hline 9 & 7 & 3 \\ \hline 8 & 6 & 2 \\ \hline \end{tabular}

Studdy Solution

STEP 1

1. The data provided is for three groups: 0 mg, 50 mg, and 100 mg.
2. We are conducting a one-way ANOVA to determine if there are statistically significant differences between the means of these three groups.
3. The test statistic F F is calculated using the ANOVA formula.

STEP 2

1. Calculate the mean for each group.
2. Calculate the overall mean of all data points.
3. Compute the between-group sum of squares (SSB).
4. Compute the within-group sum of squares (SSW).
5. Calculate the degrees of freedom for between-group and within-group.
6. Compute the mean square between (MSB) and mean square within (MSW).
7. Calculate the F-statistic.
8. Compare the calculated F-statistic to the given options.

STEP 3

Calculate the mean for each group.
- For 0 mg: $ \text{Mean}_{0\text{mg}} = \frac{9 + 8 + 7 + 8 + 8 + 9 + 8}{7} = 8.14 \]
- For 50 mg: $ \text{Mean}_{50\text{mg}} = \frac{7 + 6 + 6 + 7 + 8 + 7 + 6}{7} = 6.71 \]
- For 100 mg: $ \text{Mean}_{100\text{mg}} = \frac{4 + 3 + 2 + 3 + 4 + 3 + 2}{7} = 3.00 \]

STEP 4

Calculate the overall mean of all data points.
Overall Mean=(9+8+7+8+8+9+8)+(7+6+6+7+8+7+6)+(4+3+2+3+4+3+2)21=5.95\text{Overall Mean} = \frac{(9 + 8 + 7 + 8 + 8 + 9 + 8) + (7 + 6 + 6 + 7 + 8 + 7 + 6) + (4 + 3 + 2 + 3 + 4 + 3 + 2)}{21} = 5.95

STEP 5

Compute the between-group sum of squares (SSB).
SSB=7×((8.14−5.95)2+(6.71−5.95)2+(3.00−5.95)2)\text{SSB} = 7 \times ((8.14 - 5.95)^2 + (6.71 - 5.95)^2 + (3.00 - 5.95)^2)
=7×(4.8121+0.5776+8.7025)=7×14.0922=98.6454= 7 \times (4.8121 + 0.5776 + 8.7025) = 7 \times 14.0922 = 98.6454

STEP 6

Compute the within-group sum of squares (SSW).
SSW=∑i=17((xi,0mg−8.14)2+(xi,50mg−6.71)2+(xi,100mg−3.00)2)\text{SSW} = \sum_{i=1}^{7} ((x_{i,0\text{mg}} - 8.14)^2 + (x_{i,50\text{mg}} - 6.71)^2 + (x_{i,100\text{mg}} - 3.00)^2)
Calculate each term and sum them up:
=(0.762+0.142+1.142+0.142+0.142+0.862+0.142)+(0.292+0.712+0.712+0.292+1.292+0.292+0.712)+(1.002+0.002+1.002+0.002+1.002+0.002+1.002)= (0.76^2 + 0.14^2 + 1.14^2 + 0.14^2 + 0.14^2 + 0.86^2 + 0.14^2) + (0.29^2 + 0.71^2 + 0.71^2 + 0.29^2 + 1.29^2 + 0.29^2 + 0.71^2) + (1.00^2 + 0.00^2 + 1.00^2 + 0.00^2 + 1.00^2 + 0.00^2 + 1.00^2)
=1.2996+1.9996+5.0000=8.2992= 1.2996 + 1.9996 + 5.0000 = 8.2992

STEP 7

Calculate the degrees of freedom for between-group and within-group.
- Between-group degrees of freedom: dfB=k−1=3−1=2 df_B = k - 1 = 3 - 1 = 2 - Within-group degrees of freedom: dfW=N−k=21−3=18 df_W = N - k = 21 - 3 = 18

STEP 8

Compute the mean square between (MSB) and mean square within (MSW).
MSB=SSBdfB=98.64542=49.3227\text{MSB} = \frac{\text{SSB}}{df_B} = \frac{98.6454}{2} = 49.3227
MSW=SSWdfW=8.299218=0.4611\text{MSW} = \frac{\text{SSW}}{df_W} = \frac{8.2992}{18} = 0.4611

STEP 9

Calculate the F-statistic.
F=MSBMSW=49.32270.4611≈106.99F = \frac{\text{MSB}}{\text{MSW}} = \frac{49.3227}{0.4611} \approx 106.99

STEP 10

Compare the calculated F-statistic to the given options.
The calculated F-statistic F≈106.99 F \approx 106.99 does not match any of the given options. Therefore, there might be a calculation error or a misinterpretation of the problem. Please verify the problem statement and data.

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