Question

Read the following study description and answer the questions below. When mathematical calculations are needed, be sure to show your work and write the answer in the space below (or in your work & underline). You may use Excel, but you'll still need to show your work People who are concerned about their health may prefer non-beef hot dogs because they believe they contain fewer calories. Below are some (real) data comparing calories found in 51 brands of beef, poultry, and mixed-meat hot dogs. Do these data support the conclusion that hot dogs containing different meats differ in calorie content? 182 Beef х X-M)^2 186 672.88 181 438.48 176 254.08 149 122.32 184 573.12 190 896.4 158 4.24 139 443.52 175 223.2 148 145.44 152 64.96 2406.88 141 363.28 153 49.84 190 896.4 157 9.36 131 844.48 2721 17 160.06 8408.88 Poultry X (X-M)^2 129 104.86 132 175.3 102 280.9 106 162.82 94 613.06 102 280.9 87 1008.7 99 390.46 107 138.3 113 33.18 135 263.74 142 540.1 86 1073.22 143 587.58 152 1104.9 146 742.02 144 637.06 2019 17 118.76 8137.1 Mixed-meat х (X-M)^2 173 204.2 191 1042.64 542.42 190 979.06 172 176.62 147 137.12 146 161.54 139 388.48 175 265.36 136 515.74 179 411.68 153 32.6 107 2673.92 195 1316.96 562.16 140 350.06 138 428.9 2698 17 158.71 10189.46 135 ΣΧ- N= M= SS ΣΧ NE M= ΣΧ N- M- SS SS- 1. What is the name of the test to be performed? 2. What is the value of the test statistic for these data? 3. What is the statistical decision (with a =.05)? 4. What proportion of the variance in calories is accounted for by meat content? 5. Write an APA-style sentence describing these results.

Answer 1

The hypotheses to test are as below:

H₀: The hot dog containing different mean does not differ in calorie content i.e.µ₁ = µ₂ =µ₃

H_a: The hot dog containing different mean differs in calorie content i.e. at least two of the three µ_i’s are different

We have

	Beef	Poultry	Mixed Meat
Sample size	17	17	17
Total	2721	2019	2698
Mean	160.06	118.76	158.71
SS	8408.88	8137.1	10189.46

The overall mean of the data is

M = Xqnq + X₂ M₂ + łzna ni N3 +ng +

M 160.06 * 17 + 118.76 * 17 + 158.71* 17 17 + 17 + 17

7438 м 51

M = 145.84

Sum of squares due to treatment can be obtained as below:

Tr.S.S.= Ση, (5, - Μ): =1

Tr.S.S.= 17 * (160.06 - 145.84)+17 * (118.76 – 145.84)+17 (158.71 – 145.84)2

Tr.S.S.= 3436.026+12469.444 2814.455

Tr.S.S. = 18713.22

Sum of squares due to error (variance within) can be obtained as below:

E.S.S. = Σ SS,

E.S.S.= 8408.88 + 8137.1 + 10189.46

E.S.S. = 26735.53

Total sum of squares = 18713.22+26735.53

TSS= 45448.75

The ANOVA table can be formed as below:

Source of Variation	Degrees of Freedom	Sum of Squares	Mean Squares	F Ratio	p-value
Treatment	2	18713.22	9356.61	16.80	0000
Error	48	26735.53	556.99
Total	50	45448.75

The test statistic is F value in above ANOVA table

F=16.80

The critical value of F is table value of F at 0.05 level with (2, 48) df

F_(0.95,2,48) = 3.19

We have F statistic =16.80 which is greater than critical value of F and it indicates that we have strong evidence against null hypothesis to reject it so we reject the H₀ at 0.05 level of significance.

The proportion of variance in calories that is accounted for by meat content can be obtained as below:

T1 SS n² TSS

18713.22 ท 45448.75

m2 = 0.412

So we can say that 41.2% of variance in calories is accounted for by meat content.

Now we answer the given questions:

Answer(1):

Name of the test to be performed: One – way Analysis of variance (F test)

Answer(2):

Value of test statistic:

F= 16.80

Answer(3):

Statistical Decision: Reject the Null Hypothesis

Answer(4):

The proportion of variance in calories that is accounted for by meat content = 41.2%

Answer(5):

A one-way between subjects ANOVA was conducted to compare the effect of meat content on calorie content in Beef, Poultry, and Mixed Meat.

There was a significant effect of Meat content on calorie content at the p<.05 level for the three conditions [F(2,48) =16.80, p = 0.0000].