Question

Head movement evaluations are important because disabled individuals may be able to operate communications aids using head motion. A paper reported the accompanying data on neck rotation (in degrees) for 14 subjects both in the clockwise direction (CL) and in the counterclockwise direction (CO). For purposes of this exercise, you can assume that the 14 subjects are representative of the population of adult Americans.Subject: 1 2 3 4 5 6 7 8 9 10 11 12 13 14CL: 57.8 35.7 54.5 56.8 51.1 70.8 77.3 51.6 54.7 63.6 59.2 59.2 55.8 38.5CO: 44.7 52.1 60.2 52.7 47.2 65.6 71.4 48.8 53.1 66.3 59.8 47.5 64.5 34.4Based on these data, is it reasonable to conclude that mean neck rotation is greater in the clockwise direction than in the counterclockwise direction? Carry out a hypothesis test using a significance level of 0.01. (Use a statistical computer package to calculate the P-value. Use μCL − μCO.

Answer 1

Answer:

$t=\frac{56.186 -54.879}{\sqrt{\frac{10.813^2}{10}+\frac{10.200^2}{14}}}=0.329$

$p_v =P(Z>0.329)=0.372$

And we can calculate the p value with the following excel code:

"=1-T.DIST(0.329,26,TRUE)"

And with 1% of significance we have enough evidence to FAIL reject the null hypothesis since the $p_v > \alpha$. So we can't conclude that the true mean for  is greater in the clockwise direction than in the counterclockwise direction at the significance level provided.

Step-by-step explanation:

Data given and notation

CL: 57.8 35.7 54.5 56.8 51.1 70.8 77.3 51.6 54.7 63.6 59.2 59.2 55.8 38.5

CO: 44.7 52.1 60.2 52.7 47.2 65.6 71.4 48.8 53.1 66.3 59.8 47.5 64.5 34.4

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

We can calculate the mean and sample deviations from the data given with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X_{CL} =56.186$ represent the sample mean 1 clockwise direction

$\bar X_{CO} =54.879$ represent the sample mean 2counterclockwise direction

$n_{CL}$=14 represent the sample clockwise direction  size  

$n_{CO}$=14 represent the sample counterclockwise direction size  

$s_{CL} =10.813$ sample standard deviation for sample clockwise direction

$s_{CO} =10.200$ sample standard deviation for sample  counterclockwise direction

$\mu_1 -\mu_2$ parameter of interest.

System of hypothesis

We want to chekc if the mean neck rotation is greater in the clockwise direction than in the counterclockwise direction.

Null Hypothesis:$\mu_{CL} -\mu_{CO} \leq 0$

Alternative hypothesis:$\mu_{CL} -\mu_{CO} > 0$

Calculate the statistic

The statistic would be:

$t=\frac{\bar X_{CL} -\bar X_{CO} -0}{\sqrt{(\frac{s^2_{CL}}{n_{CL}}+\frac{s^2_{CO}}{n__{CO}})}}$

And if we replace we got:

$t=\frac{56.186 -54.879}{\sqrt{\frac{10.813^2}{10}+\frac{10.200^2}{14}}}=0.329$

Calculate the p value

First we need to calculate the degrees of freedom given by:

$df= n_{CL}+n_{CO}-2= 14+14-2=26$

And the p value would be given by:

$p_v =P(Z>0.329)=0.372$

And we can calculate the p value with the following excel code:

"=1-T.DIST(0.329,26,TRUE)"

Conclusion

And with 1% of significance we have enough evidence to FAIL reject the null hypothesis since the $p_v > \alpha$. So we can't conclude that the true mean for  is greater in the clockwise direction than in the counterclockwise direction at the significance level provided.