When 5 randomly selected people all have the same birth month of March.

Head movement evaluations are important because disabled individuals may be able to operate communications aids using head motio

earnstyle [38]

Answer:

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{1.307 -0}{\frac{7.798}{\sqrt{14}}}=0.6271$

df=n-1=14-1=13

$p_v =P(t_{(13)}>0.6271) =0.2707$

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

Let put some notation

x=test value CL , y = test value CO

x: 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

y: 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

The system of hypothesis for this case are:

Null hypothesis: $\mu_x- \mu_y \leq 0$

Alternative hypothesis: $\mu_x -\mu_y >0$

The first step is calculate the difference $d_i=x_i-y_i$ and we obtain this:

d: 13.1, -16.4,-5.7,4.1,3.9,5.2,5.9,2.8,1.6,-2.7,-0.6,11.7,-8.7,4.1

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{29}{8}=1.307$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =7.798$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{1.307 -0}{\frac{7.798}{\sqrt{14}}}=0.6271$

The next step is calculate the degrees of freedom given by:

$df=n-1=14-1=13$

Now we can calculate the p value, since we have a right tailed test the p value is given by:

$p_v =P(t_{(13)}>0.6271) =0.2707$

So the p value is higher than the significance level provided of 0.01, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is higher or equal than 0. So we can't conclude that the mean neck rotation is greater in the clockwise direction than in the counterclockwise direction at 1% of significance.

8 0

3 years ago

Community college students conduct a survey at their college. They ask "Do you plan to transfer to a university to pursue a bacc

Tamiku [17]

Answer:

$0.8 - 2.58\sqrt{\frac{0.8(1-0.8)}{100}}=0.697$

$0.8 + 2.58\sqrt{\frac{0.8(1-0.8)}{100}}=0.903$

The 99% confidence interval would be given by (0.697;0.903)

And the best option is : 0.697 to 0.903

Step-by-step explanation:

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".

Solution to the problem

For this case the conditions we have the conditions to assume a normal distribution for the population proportion since:

np=100*0.8=80 >10 , n(1-p) = 10*(1-0.8) =20 >10

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by: