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ivanzaharov [21]
3 years ago
6

What is the mode for the set of data? 6,7,10,12,12,13 10 12 13 11

Mathematics
1 answer:
KonstantinChe [14]3 years ago
4 0
12
because it is the only number that appears more than once which is basically the definition of mode.
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Listed below are systolic blood pressure measurements (mm Hg) taken from the right and left arms of the same woman. Consider the
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a) \bar d= \frac{\sum_{i=1}^n d_i}{n}=72.2  

s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =9.311  

b) The 90% confidence interval would be given by (63.330;81.070)

63.330 < \mu_{left}- \mu_{right}  

Step-by-step explanation:

1) Previous concepts  and notation

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".

Let put some notation  

x=left arm , y = right arm  

x: 175 169 182 146 144  

y: 102 101 94 79 79

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

d:  73, 68, 88, 67, 65

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}=72.2  

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}} =9.311  

2) Confidence interval

The confidence interval for the mean is given by the following formula:  

\bar d \pm t_{\alpha/2}\frac{s_d}{\sqrt{n}} (1)  

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:  

df=n-1=5-1=4  

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,4)".And we see that t_{\alpha/2}=2.13  

Now we have everything in order to replace into formula (1):  

72.2-2.13\frac{9.311}{\sqrt{5}}=63.330  

72.2+2.13\frac{9.311}{\sqrt{5}}=81.070  

So on this case the 90% confidence interval would be given by (63.330;81.070)

63.330 < \mu_{left}- \mu_{right}  

3 0
3 years ago
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