Question

Foot-pressure Angle. Genu valgum, commonly known as "knee-knock, is a condition in which the knees angle in and touch one another when standing. Genu varum, commonly known as "bow-legged," is a condition in which the knees angle out and the legs bow when standing. In the article "Frontal Plane Knee Angle Affects Dynamic Postural Control Strategy during Unilateral Stance" (Medicine and Science in Sports de Exercise, Vol. $34$, No, $7$, Pp. $1150-1157$), J. Nyland et al studied patients with and without these conditions, One aspect of the study was to see whether patients with genu valgum or genu varum had a different angle of foot pressure when standing. The following table provides summary statistics for the angle, in degrees, of the anterior-posterior center of foot pressure for patients that have genu valgum, genu varum, or neither condition.

At the significance level. do the data provide sufficient evidence to conclude that a difference exists in the mean angle of anterior-posterior center of foot pressure among people in the three condition groups? Note; For the degrees of freedom in this exercise:

Short answer

The data provided is sufficient $t$to conclude a difference in the mean angle of anterior-posterior centre foot pressure among people in the three condition groups.

Step-by-step solution

Given information

The given data is

Explanation

From the given data

The level of significance $\alpha =0.05$

Consider the test hypothesis

Null hypothesis:

$H_0$: There is no evidence that data provide is sufficient t conclude that a difference in mean angle of anterior-posterior centre $f$foot pressure among people in the three condition groups

Alternative hypothesis:

$H_a$: There is an evidence that the data provided is sufficient $t$conclude that a difference in mean angle of anterior-posterior centre $f$foot pressure among people in the three condition groups

Total number of samples

$n=\sum n$

$=11+16+29$

$=56$

The mean of all observations

$\overline{x}=\frac{{\sum }^{n_i{x}_i}}{{\sum }^{n_i}}$

$=\frac{11(60.6)+16(60.7)+29(54.2)}{56}$

$=57.314$

The treatment sum of squares

$SSTR=\sum n_i{\left(\overline{x_i}-\overline{x}\right)}^2$

$=11(60.6-57.3143{)}^2+16(60.7-57.3143{)}^2+29(54.2-57.3143{)}^2$

$=2383$

The total sum of squares

$SST=SSTR+SSE$

$=583.4286+2383$

$=2966.4286$

The mean treatment sum of squares

$MSTR=\frac{SSTR}{k-1}$

$=\frac{583.4286}{k-1}$

$=291.7143$

The mean error sum of squares

$MSE=\frac{SSE}{n-k}$

$=\frac{2383}{568-3}$

$=44.96$

The $F-$static is

$F-static=\frac{MSTR}{MSE}$

$=\frac{291.7143}{44.9623}$

$=6.488$

The one-way ANOVA table is

The critical value at $1\%$significant level is $5.03$.

Critical value approach: The level of significance $\alpha =0.01$

$F-statistic(6.488)>\alpha (0.01)$The null hypothesis is rejected

As a result, the null hypothesis is rejected at a significant level of $1\%.$

As a result, the results are statistically significant at the $1\%$level.

As a result, the data show that there is a difference in the mean angle of anterior-posterior centre $f$foot pressure between participants in the three condition groups.