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ss7ja [257]
3 years ago
11

Yvonne is interested in testing the effectiveness of a new flea medication on dogs, so she divided some dogs with fleas into two

groups, group A and group B. She gave group A the flea medication. Should she also give group B the flea medication?
Mathematics
2 answers:
meriva3 years ago
7 0

Actually, the real answer is "NO, BECAUSE GROUP B SHOULD BE THE CONTROL GROUP". Since Yvonne gave the flea medication to the group A, and we need a control group. If you don't know what it's the control group, it's the group that they don't put the experiment on, just to see if the group(s) being experimented, had any positive or negative changes. In a short summary they need group B to stay like they are to compare it to the experimented group (group A)

Good Luck!

expeople1 [14]3 years ago
7 0

Answer:

Answer is

No because group B sould be in control

Step-by-step explanation:

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Step-by-step explanation:

a)82km into miles

Sol'n

Here,

km=82=82000miles

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Step-by-step explanation:

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If a test of H subscript 0 colon space mu subscript D equals 0 space v s. space H subscript a colon space space mu subscript D g
lara [203]

Answer:

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.

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
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