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nlexa [21]
3 years ago
12

Two variables are correlated with r = -0.23. Which description best describes the strength and direction of the association betw

een the variables?
A. strong negative


B. weak positive


C. weak negative


D. strong positive
Mathematics
1 answer:
tatuchka [14]3 years ago
3 0
Answer is C weak negavite


weak, because as the value became smaller that 1 the correlation weakens. negavite because it is a negative value (-0.23)
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2 years ago
An article describes a study comparing two methods of measuring mean arterial blood pressure. The auscultatory method is based o
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p_v =P(t_{(5)}>0.499) =0.319

The p value is higher than any significance level given for example \alpha=0.05,0.1, so then we can conclude that we FAIL to reject the null hypothesis. So we don't have enough evidence to conclude that the mean reading is greater for the auscultatory method at 5% or 10% of significance.

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 we can use it.  

Let put some notation  

x=Auscultatory method , y = Oscillatory method

x: 74 86 84 79 70 78 79 70

y: 70 85 90 110 71 80 69 74

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 define the difference d_i=x_i-y_i, that is given so we have:

d: 6.6, 4.2, -5.5, -3.1, 9.3, -3.9

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}=1.267

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

s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =6.215

The fourth step is calculate the statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{1.267 -0}{\frac{6.215}{\sqrt{6}}}=0.499

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

df=n-1=6-1=5

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

p_v =P(t_{(5)}>0.499) =0.319

The p value is higher than any significance level given for example \alpha=0.05,0.1, so then we can conclude that we FAIL to reject the null hypothesis. So we don't have enough evidence to conclude that the mean reading is greater for the auscultatory method at 5% or 10% of significance.

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