Stress levels and viral infections: an ANOVA analysis

Statistics and Probability

A researcher investigated the number of viral infections people contract as a function of the amount of stress they experienced during a 6-month period. The obtained data:

Amount of Stress

Negligible Stress

Minimal Stress

Moderate Stress

Severe Stress

What are Ho and Ha?

Ho: Stress levels experienced during a six-month period does not have a significant impact on viral infection rates.

HA: Stress levels do have a significant impact on a person's susceptibility to acquiring viral infections over time.

Compute Fobt and complete the ANOVA summary table.

Total Sum of Squares:

Xtot = 2+4+6+5+1+ . . . +8+1+3+5+4 = 67

(?Xtot) 2 = 672 = 4489

X2tot = 4+16+36+25+1+ . . . +64+1+9+25+16 = 349

ntot = 16

SStot = ?X2tot -- [(?Xtot) 2/ntot] = 349 -- [4489/16] = 349 -- 280.6 = 68.4

Within Groups Sum of Squares:

Xno = 2+1+4+1 = 8

(?Xno) 2 = 82 = 64

X2no = 4+1+16+1 = 22

nno = 4

SSno = ?X2no -- [(?Xno) 2/nno] = 22 -- [64/4] = 6

Xmin = 4+3+2+3 = 12

(?Xmin) 2 = 122 = 144

X2min = 32+16+9+4+9 = 38

nmin = 4

SSmin = ?X2min -- [(?Xmin) 2/nmin] = 38 -- [144/4] = 2

Xmod = 6+5+7+5 = 23

(?Xmod) 2 = 232 = 529

X2mod = 36+25+49+25 = 135

nmod = 4

SSmod = ?X2mod -- [(?Xmod) 2/nmod] = 135 -- [529/4] = 2.75

Xsevere = 5+7+8+4 = 24

(?Xsevere) 2 = 242 = 576

X2severe = 25+49+64+16 = 154

nsevere = 4

SSsevere = ?X2severe -- [(?Xsevere) 2/nsevere] = 154 -- [576/4] = 10

SSwithin = ?X2no -- [(?X no) 2/n no] + . . . + ?X2severe -- [(?Xsevere) 2/nsevere] =

6+2+2.75+10 = 20.75

Between Groups Sum of Squares:

SSbetween = [(?Xno) 2/nno)]+ [(?XMmin) 2/nmin)]+ [(?Xmod) 2/nmod)]+ [(?Xsevere) 2/nsevere)] -- [(?Xtot) 2/ntot)] = (64/4)+(144/4)+(529/4)+(576/4) -- (4489/16) =

16+36+132.25+144 -- 280.6 = 47.69

Degrees of Freedom:

dftot = ntot -- 1 = 16 -- 1 = 15

dfwithin = ntot -- #groups = 16 -- 4 = 12

dfbetween = #groups -- 1 = 4 -- 1 = 3

Mean Squares:

MSbetween = SSbetween/dfbetween = 47.69/3 = 15.90

MSwithin = SSwithin/dfwithin = 20.75/12 = 1.73

F-ratio:

F = MSbetween/MSwithin = 15.90/1.73 = 9.19

ANOVA Summary Table:

Source

SS

df

MS

F

Between

47.69

3

15.90

9.19

Within

20.75

12

1.73

Total

68.44

15

C. With ? = .05, what is the Fcrit ?

Based on the F. table, using 3 degrees of freedom for the numerator and 12 degrees of freedom for the denominator, and an ? = .05, then Fcrit = 3.49.

D. Report your statistical results.

There was a significant main effect for treatment, F (3, 12) = 9.19, p < .010; therefore, the null hypothesis must be rejected.

E. Perform the appropriate post hoc comparisons.

The formula for Tukey's HSD (honestly significant difference) post-hoc test is: HSD = q*?(MSwithin/ngroup), where k = the number of groups (4) and dfwithin = 12, therefore, q = 4.20.

HSD = 4.20*?(1.73/4) = 4.20*0.658 = 2.76.

Based on the HSD value, any difference between group means greater than 2.76 would be considered significant:

Mno = 2; Mmin = 3; Mmod = 5.75; Msevere = 6

Mno vs. Mmin = 1, no significant difference since 1 < 2.76

Mno vs. Mmod = 3.75, significant difference since 3.75 > 2.76

Mmin vs. Mmod = 2.75, no significant difference since 2.75 < 2.76

Mmin vs. Msevere = 3.0, significant difference since 3.0 > 2.76

Mmod vs. Msevere = 0.25, no significant difference since 0.25 < 2.76

F. What do you conclude about this study?

The one-way ANOVA revealed a significant difference between the groups. Post-hoc analysis using Tukey's HSD test revealed that the difference between the negligible stress group and the moderate and severe stress groups were significant. In addition, the difference between the minimum and severe stress groups was also significant. This finding suggests that there is significant relationship between stress levels and susceptibility to viral infections, although the direction of causality is unknown.

G. Compute the effect size and interpret it.

2 = [SSbetween -- (k -- 1)(MSwithin)]/(SStot + MSwithin), where k equals number of groups

2 = [47.69 -- (4 -- 1)(1.73)]/(68.4+1.73) = 42.5/70.13 = 0.606 = 60.6%

This finding suggests that 60.6% of the variability in viral infections between groups can be explained by stress levels.

H. Estimate the value of µ that is likely to be found in the severe stress condition.

µ = ?X/N = 24/4 = 6

Question 2

In a study in which k = 3, n = 21, X1 = 45.3, X2 = 16.9, X3 = 8.2 compute the following squares.

Source

Sum of Squares

df

Mean Square

F

Between

2

73.66

1.54

Within

18

47.94

Total

20

A. Complete the ANOVA summary table.

Degrees of Freedom:

dftot = ntot -- 1 = 21 -- 1 = 20

dfwithin = ntot -- #groups = 21 -- k = 21 -- 3 = 18

dfbetween = #groups -- 1 = k -- 1 = 3 -- 1 = 2

Mean Squares:

MSbetween = SSbetween/dfbetween = 147.32/2 = 73.66

MSwithin = SSwithin/dfwithin = 862.99/18 = 47.94

F-ratio:

F = MSbetween/MSwithin = 73.66/47.94 = 1.54

B. With ? = .05, what do you conclude about Fobt ?

Using the F. statistic table, with a 2 in the numerator and 18 in the denominator, Fcrit is 3.55 for ? Of 0.05. Since Fobt < Fcrit, the difference between the groups is nonsignificant.

C. Report your results in the correct format.

The treatment did not have a significant effect, F (2, 18) = 1.54, p > .10; therefore, the null hypothesis cannot be rejected.

D. Perform the appropriate post hoc comparisons. What do you conclude about this relationship?

Given HSD = q*?(MSwithin/ngroup), where k = 3 and dfwithin = 18, q = 3.61 for ? = .05.

HSD = 3.61*?(47.94/7) = 3.61*2.62 = 9.46.

Based on the HSD value, any difference between group means greater than 9.46 would be considered significant.

M1 = 45.3; M2 = 16.9; M3 = 8.2

M1 vs. M2 = 28.4, significant difference since 28.4 > 9.46

M1 vs. M3 = 37.1, significant difference since 37.1 > 9.46

M2 vs. M3 = 8.7, no significant difference since 8.7 < 9.46

Based on the post-hoc analysis there is a significant difference between the group 1 mean and the other group means, but no difference between the means for group 2 and 3.

E. What is the effect size in this study, and what does this tell you about the influence of the independent variable?

2 = [SSbetween -- (k -- 1)(MSwithin)]/(SStot + MSwithin)

2 = [147.32 -- (3 -- 1)(47.94)]/(1010.31+47.94) = 95.88/1058.25 = 0.0906 = 9.06%

This finding suggests the independent variable is responsible for 9.06% of the variability of the dependent variable. The independent variable is therefore a poor predictor of X.

Question 3

A researcher investigated the effect of volume of background noise on participant's accuracy rates while performing a difficult task. He tested three groups of randomly selected students and obtained the following means and sums of squares:

Low Volume

Moderate Volume

High Volume

X

61.5

65.5

48.25

n

4

5

7

Source

Sum of Squares

df

Mean Square

F

Between Groups

2

6.92

Within Groups

13

47.09

Total

15

A. Complete the ANOVA.

Degrees of Freedom:

dftot = ntot -- 1 = 16 -- 1 = 15

dfwithin = ntot -- #groups = 16 -- 3 = 13

dfbetween = #groups -- 1 = 3 -- 1 = 2

Mean Squares:

MSbetween = SSbetween/dfbetween = 652.16/2 = 326.08

MSwithin = SSwithin/dfwithin = 612.16/13 = 47.09

F-ratio:

F = MSbetween/MSwithin = 326.08/47.09 = 6.92

B. At ? = .05, what of Fcrit?

Using ? = .05, a numerator of 2, and a denominator of 13, the Fcrit value equals 3.81.

C. Report the statistical results in the proper format.

There was a significant main effect for treatment, F (2, 13) = 6.92, p < .010; therefore, the null hypothesis must be rejected.

D. Perform the appropriate post hoc tests.

Using the Tukey-Kramer post-hoc test for unequal group sample sizes:

HSD = q*?[(MSwithin/2)(1/n1+1/n2)], where k = 3, dfwithin = 13, q = 3.73, for ? = .05.

HSDlow vs. mod = 3.73*?[(47.09/2)(1/4+1/5)] = 12.14

HSDlow vs. high = 3.73*?[(47.09/2)(1/4+1/7)] = 11.34

HSDmod vs. high = 3.73*?[(47.09/2)(1/5+1/7)] = 10.60

Mlow = 61.5; Mmod = 65.5; Mhigh = 48.25

Mlow vs. Mmod = 4.0, no significant difference since 4.0 < 12.14

Mlow vs. Mhigh = 13.25, significant difference since 13.25 > 11.34

Mmod vs. Mhigh = 17.25, significant difference since 17.25 > 10.60

E. What do you conclude about this study?

Based on the post-hoc analysis there is a significant performance accuracy difference between the group exposed to high volume background noise and participants exposed to low and moderate levels of background noise.

F. Compute the effect size and interpret it.

2 = [SSbetween -- (k -- 1)(MSwithin)]/(SStot + MSwithin), where k equals number of groups

2 = [652.16 -- (3 -- 1)(47.09)]/(1264.92+47.09) = 557.98/1312.01 = 0.425 = 42.5%

This finding suggests that 42.5% of the variability in performance accuracy is associated with the levels of background noise.

Question 4

In an experiment, you measure the popularity of two brands of soft drinks (factor A) and for each brand you test males and females (factor B). The following table shows the main effect and cell means from the study:

Factor A

Level A1:

Brand X

Level A2:

Brand Y

Factor B

Level B1:

Males

14

23

Level B2:

Females

25

12

A. Describe the graph of the interaction when factor A is on the X axis.

Using a line plot with factor A on the X axis, the line for males has an upward slope from brand X to brand Y, while the line representing females has a downward slope.

Question 5

You conduct an experiment involving two levels of self-confidence ( A1 is low and A2 is high) and examine participants' anxiety scores after they speak to one of four groups of differing sizes ( B1 through B4 represent speaking to a small, medium, large, or extremely large group, respectively). You compute the following sums of squares (n = 4 and N = 32):

Source

Sum of Squares

df

Mean Square

F

Between

Factor A

8.42

1

8.42

2.20

Factor B

76.79

3

25.60

6.70

Interaction

23.71

3

7.90

2.07

Within

29

3.82

Total

31

7.09

A. Complete the ANOVA summary table.

The independent variables are self-confidence (factor A) and group size (factor B), while the dependent variable is anxiety scores.

a = levels of self-confidence = 2

c = audience sizes = 4

ac = cells in experiment = 8

N = total number anxiety scores = 32

n = number of anxiety scores per cell = 4

dfa = a -- 1 = 2 -- 1 = 1

dfc = c -- 1 = 4 -- 1 = 3

dfac = (a -- 1)(c -- 1) = 1*3 = 3

dfwithin = N -- ac = 32 -- 3 = 29

dftot = N -- 1 = 31

MSa = SSa/dfa = 8.42/1 = 8.42

MSc = SSc/dfc = 76.79/3 = 25.60

MSac = SSac/dfac = 23.71/3 = 7.90

MSwithin = SSwithin/dfwithin = 110.72/29 = 3.82

MStot = SStot/dftot = 219.64/31 = 7.09

Fa = MSa/MSwithin = 8.42/3.82 = 2.20

Fc = MSc/MSwithin = 25.60/3.82 = 6.70

Fac = MSac/MSwithin = 7.90/3.82 = 2.07

B. With a ? = .05, what do you conclude about each Fobt?

From the F. table: Fcrit (1, 29) and Fcrit (3, 29) are 4.18 and 2.93, respectively.

Since Fa (1, 29) < Fcrit (1, 29), then there is no main effect of levels of self-confidence on anxiety scores; therefore, the null hypothesis (Ho: µlow = µhigh) must be retained.

Since Fc (3, 29) > Fcrit (3, 29), there is a main effect of audience size on anxiety scores and the null hypothesis (Ho: µS = µM = µL = µEL) must be rejected.

Since Fac (3, 29) < Fcrit (3, 29), then there is no significant interaction between levels of self-confidence and audience size; therefore, the null hypothesis [Ho: (µlow-S -- µhigh-S) = (µlow-M -- µhigh-M) = (µlow-L -- µhigh-L) = (µlow-EL -- µhigh-EL)] must be retained.

C. Compute the appropriate values of HSD.

Given k = 8 and dfwithin = 29, then q = 4.60

HSD = 4.60*?(3.82/8) = 4.60*0.691 = 3.18

Based on the HSD value, any difference between audience-size group means greater than 3.18 would be considered significant:

D. For the levels of factor B, the means are X1 = 18.36, X2 = 20.02, X3 = 24.6, and X4 = 28.3. What should you conclude about the main effect of B?

X1 vs. X2 = |18.36 -- 20.02| = 1.66; since 1.66 < 3.18 then there is no significant difference in anxiety scores between the small and medium sized audiences.

X1 vs. X3 = |18.36 -- 24.6| = 6.24; since 6.24 > 3.18, the anxiety scores for the small audience size is significantly different from the anxiety scores for both the large and extremely large audience sizes.

X2 vs. X3 = |20.02 -- 24.6| = 5.54; since 5.54 > 3.18, the anxiety scores for the medium sized audience is significantly different from the anxiety scores for both the large and extremely large audience sizes.

X3 vs. X4 = |24.6 -- 28.3| = 3.70; since 3.70 > 3.18, the anxiety scores for the large and extremely large audience sizes are significantly different.

E. How important is the size of the audience in determining a person's anxiety score> How important is the persons self-confidence?

2 = [SSbetween -- ([df)(MSwithin)]/(SStot + MSwithin)

a2 = [8.42 -- (1)(3.82)]/(219.64+3.82) = 4.60/223.42 = 0.021 = 2.1%

c2 = [76.79 -- (3)(3.82)]/(219.64+3.82) = 65.33/223.42 = 0.292 = 29.2%

Audience size explains close to 30% of the variability in anxiety scores between groups, while a person's self-confidence only explains 2.1%. Audience size is therefore close to 15 times more important than self-confidence in determining anxiety scores.