Question

Section Exercise 11-8 Sales of People magazine are compared over a 5-week period at four Borders outlets in Chicago. Weekly Sales Store 1 Store 2 Store 3 Store 4 102 97 89 100 106 77 91 116 105 82 75 87 115 80 106 102 112 101 94 100 Click here for the Excel Data File Fill in the missing data. (Round your p-value to 4 decimal places, mean values to 1 decimal place, and other answers to 2 decimal places.) Treatment Mean n Std. Dev Store 1 Store 2 Store 3 Store 4 Total One-Factor ANOVA Source SS df MS F p-value Treatment Error Total (a) Based on the given hypotheses choose the correct option. H0: μ1 = μ2 = μ3 = μ4 H1: Not all the means are equal α = 0.05 Reject the null hypothesis if F > 3.24 Reject the null hypothesis if F < 3.24 (b) Determine the value of F. (Round your answer to 2 decimal places.) F-value (c) On the basis of the above-determined values, choose the correct decision from below. Fail to reject the null hypothesis. Reject the null hypothesis. (d) Determine the p-value. (Round your answer to 4 decimal places.) p-value Next Visit question mapQuestion 1 of 2 Total1 of 2 Prev

Answer 1

Answer:

Step-by-step explanation:

Hello!

An ANOVA was conducted to analyze the variable

Y: sales of "People" magazine over a 5-week period

This was studied in 4 Borders outlets in Chicago.

So this test has one factor: "Borders outlets" and four treatments: "Store 1, store 2, store 3 and store 4"

For each store you have the data for the weekly sales over a 5-week period so the sample sizes are:

n₁=n₂=n₃=n₄= 5 weeks

Store 1

∑X₁= 540; ∑X₁²= 58434

X[bar]₁= 108

S₁²= 28.50

S₁= 5.34

Store 2

∑X₂= 437; ∑X₂²= 38663

X[bar]₂= 87.40

S₂²= 117.30

S₂= 10.83

Store 3

∑X₃= 455; ∑X₃²= 41899

X[bar]₃= 91

S₃²= 123.50

S₃= 11.11

Store 4

∑X₄=505; ∑X₄²= 51429

X[bar]₄= 101

S₄²= 106

S₄= 10.30

Totals

N= n₁ + n₂ + n₃ + n₄= 4*5= 20

∑Mean= 108+87.40+91+101= 387.4

∑Variance= 28.50+117.30+123.50+106= 375.30

∑Standard deviation= 5.34+10.83+11.11+10.30= 37.58

Hypothesis test:

H₀: μ₁= μ₂= μ₃= μ₄

H₁: At least one population mean is different.

α: 0.05

The statistic for this test is

$F= \frac{MS_{Treatments}}{MS_{Error}} ~~F_{k-1;N-k}$

k-1= 3 Df of the treatments, k=4 number of treatments

N-k= 16 Df of errors, N=20 total number of observations in all treatments

$F_{H_0}= \frac{441.78}{93.83}= 4.71$

The critical region and p-value for this test are one-tailed to the right.

Using the critical value approach:

$F_{k-1;N-k;1-\alpha }= F_{3;16;0.95}= 3.25$

The decision rule is:

If $F_{H_0}$ ≥ 3.25, reject the null hypothesis.

If $F_{H_0}$ < 3.25, do not reject the null hypothesis.

Using the p-value approach:

Little reminder: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

So under the distribution F₃,₁₆ you have to calculate the probability of the calculated  $F_{H_0}$:

P(F₃,₁₆≥4.71)= 1 - P(F₃,₁₆<4.71)= 1 - 0.9847 = 0.0153

p-value= 0.0153

The decision rule for this approach is

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

The p-value is less than the level of significance, so the decision is to reject the null hypothesis.

I hope this helps!