Question

Asked May 21, 2019

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Listed below are amounts of arsenic in samples of brown rice from three different states. The amounts are in micrograms of arsenic and all samples have the same serving size. The data are from the Food and Drug Administration. Use a 0.05 significance level to test the claim that the three samples are from populations with the same mean. Assume that the distributions for each state are normal and that all variances are equal.

Arkansas 4.8 4.9 5.0 5.4 5.4 5.4 5.6 5.6 5.6 5.9 6.0 6.1

California. 1.5 3.7 4.0 4.5 4.9 5.1 5.3 5.4 5.4 5.5 5.6 5.6

Texas. 5.6 5.8 6.6 6.9. 6.9 6.9 7.1 7.3 7.5 7.6 7.7 7.7

Step 1

**Solution:**

**Null and alternative hypotheses:**

Null hypothesis: µ_{1 }= µ_{2} = µ_{3}

Alternative hypothesis: µ_{1} ≠ µ_{2} ≠ µ_{3}

**Test statistic:**

Step-by-step procedure to compute one-way ANOVA in EXCEL is as follows.

- Go to Data -> Data Analysis.
- Select “Anova: Single Factor” from
**Analysis Tool**. - Select three columns of data in
**Input range**and click Ok.

The output obtained as follows.

Step 2

**Decision rule:**

From the EXCEL output, P-value is 0.000 (rounded to three decimal places).

Since, P-value < 0.05, reje...

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