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ahrayia [7]
2 years ago
12

Persons having Raynaud's syndrome are apt to suffer a sudden impairment of blood circulation in fingers and toes. In an experime

nt to study the extent of this impairment, each subject immersed a forefinger in water and the resulting heat output (cal/cm2/min) was measured. For m = 9 subjects with the syndrome, the average heat output was x = 0.63, and for n = 9 nonsufferers, the average output was 2.09. Let μ1 and μ2 denote the true average heat outputs for the sufferers and nonsufferers, respectively. Assume that the two distributions of heat output are normal with σ1 = 0.3 and σ2 = 0.5.
Mathematics
1 answer:
Citrus2011 [14]2 years ago
5 0

Answer:

There is enough evidence to say that the true average heat output of persons with the syndrmoe differs from the true average heat output of non-sufferers.

Step-by-step explanation:

We have to perform a hypothesis test on the difference between means.

The null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a: \mu_1-\mu_2

μ1: mean heat output for subjects with the syndrome.

μ2: mean heat output for non-sufferers.

We will use a significance level of 0.05.

The difference between sample means is:

M_d=\bar x_1-\bar x_2=0.63-2.09=-1.46

The standard error is

s_{M_d}=\sqrt{\sigma_1^2/n_1+\sigma_2^2/n_2}=\sqrt{0.3^2/9+0.5^2/9}=\sqrt{ 0.038 } \\\\ s_{M_d}=0.194

The t-statistic is

t=\dfrac{M_d}{s_{M_d}}=\dfrac{-1.46}{0.194}=-7.52

The degrees of freedom are

df=n_1+n_2-2=9+9-2=16

The critical value for a left tailed test at a significance level of 0.05 and 16 degrees of freedom is t=-1.746.

The t-statistic is below the critical value, so it lies in the rejection region.

The null hypothesis is rejected.

There is enough evidence to say that the true average heat output of persons with the syndrmoe differs from the true average heat output of non-sufferers.

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Which of the following equations represents the area of a sector?
Rina8888 [55]

From the given options we can say that the only one that represents the area of the sector is; A = n/360 * πr²

<h3>What is the Area of the Sector?</h3>

In circles, a sector is said to be a part of a circle made of the arc of the circle together with its two radii. This means that it is a portion of the circle formed by a portion of the circumference (arc) and radii of the circle at both endpoints of the arc.

The formula for Area of a sector is given as;

θ/360 * πr²

where;

θ is the central angle of the sector

r is radius

Now, looking at the given options we can say that the only one that represents the area of the sector is;

A = n/360 * πr²

where n is the central angle of the sector

Read more about Area of Sector at; brainly.com/question/16736105

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4 0
1 year ago
Consider a t distribution with 7 degrees of freedom. Compute P(-1.29 &lt; t &lt; 1.29). Round your answer to at least three deci
ss7ja [257]

Answer:

a) 0.76197086

b) -1.73406361

Step-by-step explanation:

a)

Consider a t distribution with 7 degrees of freedom. Compute P(-1.29 < t < 1.29)

P(-1.29 < t < 1.29) would be the area under the t distribution curve with 7 degrees of freedom between -1.29 and 1.29, that is in the interval (-1.29, 1.29).

This can be done the old style by looking up in a table or by using the technology with a spreadsheet.

In Excel, the function TDIST(x,n,2) with x>0 gives the area outside the interval (-x, x) of the t distribution with n degrees of freedom.

So TDIST(1.29,7,2) gives the area outside (-1.29, 1.29).

If we subtract this value from 1 we get the desired result

Hence  

P(-1.29 < t < 1.29) = 1 - TDIST(1.29,7,2) = 1 - 0.23802914 = 0.76197086

In OpenOffice Calc, the function is the same replacing “,” with “;”  

That is

P(-1.29 < t < 1.29) = 1 - TDIST(1.29;7;2) = 0.76197086

b)

Consider a t distribution with 18 degrees of freedom. Find the value of c such that P(t≤ c) = 0.05

We are looking for a point c such that the area of the t distribution with 18 degrees of freedom to the left of c is 0.05

In Excel, the inverse function of TDIST is TINV.  

TINV(p*2,n) with p>0 gives the point c such that the area of the t distribution with n degrees of freedom to the right of c is p.  

Since <em>the t distribution is symmetric with respect to 0</em>, -c would be a point such that the area to the left of -c is p.

So we want to compute  in Excel

-TINV(0.05*2,18) = -1.73406361

In OpenOffice Calc  

-TINV(0.05*2;18) = -1.73406361

3 0
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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Children = 30

30 / 54 = 5 / 9

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