PLS HELP DUE SOON FIRST CORRECT ANSWER GETS BRAINLIEST NO LINKS

If we compare the p value with a significance level assumed $\alpha=0.05$ , we see that $p_v > \alpha$ and we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is less or equal than 0.

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 before , y = test value after

The system of hypothesis for this case are:

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

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

The first step is calculate the difference $d_i=y_i-x_i$

The second step is calculate the mean difference

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

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

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

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=t_{calculated}$

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

$df=n-1$

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

$p_v =P(t_{(n-1)}>t_{calculated}) =0.0601$

The p value on this case is given by the problem.

If we compare the p value with a significance level assumed $\alpha=0.05$ , we see that $p_v > \alpha$ and we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is less or equal than 0.

4 0

3 years ago

In your university nursing program the ratio of students who scored a grade of A vs B vs C on the final exam is 12:29:9. Assumin

Scilla [17]

Answer:

Hence, the correct option is 82%.

Step-by-step explanation:

Consider the provided information.

In your university nursing program the ratio of students who scored a grade of A vs B vs C on the final exam is 12:29:9.

We need to find the percentage of student who scored a grade of "B" or higher.

Let x is the common factor for the proportion.

Then, the number of students who got grade A is 12x

The number of students who got grade B is 29x

The number of students who got grade C is 9x.

Thus the total number of students = 12x+29x+9x = 50x

The percentage of student who scored a grade of "B" or higher:

12x+29x = 41x

$\frac{41x}{50x}\times{100}=82\%$

Hence, the correct option is 82%.

8 0

3 years ago