Question

Question Help Fair stain removal The table to the night shows the cost per ounce (in dollars) for a random sample of toothpastes exhibiting very good stain removal good stain removal and tar stain removal Ale=0.05, can you conclude that the mean costs per ounce are different? Perform a one-way ANOVA test by completing parts a through d Assume that each sample is drawn from a normal population, that the samples are independent of each other and that the populations have the same variances Very good stain removal 064 0.50 051 1.48 0.36 049 Good stain removal 022 273 034 050 0.00 0.75 1.36 (a) Identify the daim and state Ho and H. Choose the correct answer below ОА не, а да ви е OB. Ho H:At least two means are different from the others (dam) Alleast one mean is different from the others (dam) OC Hor: == (claim) OD. Ho Alleast one mean is different from the others. (claim) H. Atleast one mean is different from the others H, P. Thứ 2 - 4 - 6 (b) Identity the degrees of freedom for the numerator and for the denominator determine the critical value and determine the rejection region The degrees of freedom for the numerator di and the degrees of freedom for the denominator, dl, so The critical value is F,- (Round to two decimal places as needed) The rejection region is (Round to two decimal places as needed) (c) Calculate the test static F-D Round to three decimal places as needed) (d) Decide to reject or fail to reject the null hypothesis and interpret the decision in the content of the original claim Choose the correct deosion below O A. Since Fis not in the rejection region, fad to rejed OB. Since Fis in the rejection region, reject Hy OC. Since F is not in the rejection region, reject Hy OD. Since Fis in the rejection region, fail to reject Hy Choose the correct interpretation below. There enough evidence at the significance level to the claim that the mean cost per Ounce is Click to select your answer(s)

Answer 1

• Null hypothesis (H_o) : average cost per ounce of toothpaste exhibiting different stain removal property are same i.e., μ₁ = μ₂ = μ_3.
• Alternative Hypothesis (H₁) : Atleast one mean is different from the other (claim)

for computation of anova in spss follow the given steps:

1. Firstly enter the data into spss in the one column named, cost.
2. Now in the second column named, factor, type the values 1, 2, 3 corresponding to very good, good and fair data.
3. Now go to Analyze > Compare means > One-way Anova. Then a dialoque box will appear.
4. Then in Dependent list move cost column and in Factor move factor column.
5. Then finally click OK

X *Untitled1 [DataSeto] - IBM SPSS Statistics Data Editor File Edit View Data Transform Analyze Direct Marketing Graphs Utilities Add-ons Window Help WIVE IVE EN M EL L HA ABC A 14 HH EE HH Visible: 2 of 2 Variables Cost Factor var var var var var var var var var var var var var var 1 .64 2 -50 3 X ta One-Way ANOVA 4 .51 1.48 36 5 Dependent List: Cost Contrasts. 6 49 7 42 Post Hoc. Options... Bootstrap... 8 1.00 1.00 1.00 1.00 1.00 1.00 2.00 2.00 2.00 2.00 2.00 2.00 3.00 3.00 3.00 9 10 2.73 .58 .75 33 1.36 .34 1.31 .30 11 12 Factor: Factor 13 14 OK Paste Reset Cancel Help 15 16 17 18 19 20 21 22 23 Data View Variable View IBM SPSS Statistics Processor is ready

OUTPUT:

ANOVA Cost df F Sum of Squares 492 5.636 6.128 Mean Square .246 Sig. .605 2 Between Groups Within Groups Total 523 12 .470 14

(b) As we see in the output

1. the degrees of freedom for the numerator is 2
2. the degrees of freedom for the denominator is 12

The critical value corresponding to (2, 12) degrees of freedom at 0.05 level of significance by using f-distribution table is 3.885.

The rejection region is F > 3.885.

(c) By using ANOVA output table our test statistics is 0.523

(d). We fail to reject our null hypothesis because F is not in rejection region. Option A is correct.

There is not enough evidence at the 5% significance level to support the claim that mean cost per ounce is different.