Question

? @ * 40%( 6:43 PM uli AT&T 2 of 3 Done Part 2. Calculation test whether animal subjects consume the same amounts of sweet-tasting solutions, a esearcher has subjects consume one of three sweet-tasting solutions (sucrose, saccha- rine, or po?ycose). The amount consumed (in milliters) of each solution is given in the ta- ble. Type of Swoet Taste Solutiorn Saccharin Sucrose Polycose 3093 13 14 15 G 291 N=30 10 13 10 13 10 12 10 12 15 100 10 10 50 114 10 11.4 90.4 10 60.1

Answer 1

We need to compare more than 2 population means so one way ANOVA analysis will be used.

Hypotheses are:

H0: The consumptions of sweet solution is same for three groups.

Ha: The consumptions of sweet solution is not same for three groups.

Let G1 shows the Sucrose group, G2 shows the Saccharin group and G3 shows the Polycose group.

Here we have

$\sum G_{1}=100,\sum G_{2}=77,\sum G_{3}=114,n_{1}=10,n_{2}=10,n_{3}=10,\sum G =291,\sum G^{2}=3093$

Now

$SS_{T}=\sum G^{2}-\frac{\left (\sum G \right )^{2}}{N}=270.3$

Now

$SS_{Between}=\frac{\left (\sum G_{1} \right )^{2}}{n_{1}}+\frac{\left (\sum G_{2} \right )^{2}}{n_{2}}+\frac{\left (\sum G_{3} \right )^{2}}{n_{3}}-\frac{\left (\sum G \right )^{2}}{N} =69.8$

Now

$SS_{within}=SS_{T}-SS_{Between}=200.5$

Since there are 3 different groups so we have k=3. Therefore degree of freedoms are:

$df_{between}=k-1=3-1=2$

$df_{within}=N-k=30-3=27$

$df_{total}=27+2=29$

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Now

$MS_{between}=\frac{SS_{between}}{df_{between}}=34.9$

$MS_{within}=\frac{SS_{within}}{df_{within}}=7.425925926$

F test statistics is

$F=\frac{MS_{between}}{MS_{within}}=4.70$

So p-value of the test is 0.0177. Since P-value is less than 0.05 so we reject the null hypothesis.