Answer:
0.153
Step-by-step explanation:
We have been given that Mafalda has a save percentage of 0.738 so far this hockey season. Last season, her save percentage was 0.585.
To find improvement in Mafalda's save percentage we will subtract last season's save percentage from this season's save percentage.
Therefore, Mafalda's save percentage improved by 0.153.
Answer:
d =(m-q)/a
Step-by-step explanation:
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Answer:
Step 1: The null and alternative hypothesis for the one way analysis of variance is,
H₀ : There is no difference in the means of each area.
H₁ : At least one mean is different.
The objective of this question is to test whether the income of the three areas are significantly different.
Step 2: The problem states to use 0.05 significance level.
Step 3: The test statistic follows the F-distribution.
Step 4: Decision rule is based on the critical value.
Numerator degrees of freedom is,
K - 1 = 3 - 1
= 2
Denominator degrees of freedom is 2.
n - k = 12 - 3
= 9
Using the F-distribution table for ∝=0.05 with numerator and denominator degrees freedom, the critical value is 4.26. The decision rule is to reject H₀ if the computed value of F exceeds 4.26.
Step 5: Calculations.
Using Excel follow the steps mentioned below.
1. Import or type the data into spreadsheet.
2. Data--------> Data Analysis----------> Anova: Single Factor, click OK.
3. Select Input Range, make a tick mark on Labels in the first row, enter 0.05 for alfa.
4. Click OK.
The obtained output for the one-way ANOVA is,
[ find the figure in attachment]
From the obtained output, the computed test statistic value, F = 14.18 , which is greater than the critical value of 4.26. So, reject H₀
Step 6: Interpretation. The average income of each area is different. It can be concluded that the average income is significantly different .
