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OlgaM077 [116]
4 years ago
15

A one­sided test of a hypothesis, based on a sample of size 9, yields a P­value of .035. Which of the following best describes t

he possible range of t values that yields this P­value? .706 < t < .889 1.11 < t < 1.40 1.34 < t < 1.44 2.45 < t < 2.90 1.86 < t < 2.31
Mathematics
1 answer:
lana [24]4 years ago
8 0

Answer:

1.86 < t < 2.31

Step-by-step explanation:

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean height actually is higher\lower than an specified value, the system of hypothesis would be:  

Null hypothesis:\mu \leq \mu_o  

Alternative hypothesis:\mu > \mu_o

Or like this:

Null hypothesis:\mu \geq \mu_o  

Alternative hypothesis:\mu < \mu_o

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We don't know on this case the calculated value and with the p value we can find it.

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=9-1=8  

Since is a one side test the p value would be:  

p_v =P(t_{(8)}>t_o)=0.035

And we can find the critical value with the following excel code:

" =T.INV(1-0.035,8)" and we got t_p =2.09

And cannot be a lower tail test since all the options have positive values for the statistic. So on this case the best option is:

1.86 < t < 2.31

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From principle of floatation:

W = U

The volume of water displaced is as a result of the volume of object submerged. 

The wood would also float with 1/4 of its total height or total volume below the water surface.
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4 years ago
Examination of a large sample of people revealed a relationship between calcium intake and blood pressure. The relationship was
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Answer:

a) The 95% confidence interval for the mean is (-1.171, 11.717).

b) No, there is not enough evidence to support the claim that increasing the amount of calcium in our diet reduce blood pressure (at a 5% significance level).

Step-by-step explanation:

The data for both groups are:

Group 1 (calcium): 7 -4 18 17 -3 -5 1 10 11 -2

Group 2 (placebo): -1 12 -1 -3 3 -5 5 2 -11 -1 -3

The mean and standard deviation for Group 1 is:

M_1=\dfrac{1}{10}\sum_{i=1}^{10}(7+(-4)+18+17+(-3)+(-5)+1+10+11+(-2))\\\\\\ M_1=\dfrac{50}{10}=5

s_1=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{10}(x_i-M)^2}\\\\\\

s_1=\sqrt{\dfrac{1}{9}\cdot [(7-(5))^2+(-4-(5))^2+...+(-2-(5))^2]}\\\\\\

s_1=\sqrt{\dfrac{1}{9}\cdot [(4)+(81)+(169)+(144)+(64)+(100)+(16)+(25)+(36)+(49)]}\\\\\\s_1=\sqrt{\dfrac{688}{9}}=\sqrt{76.44}\\\\\\s_1=8.743

The mean and standard deviation for Group 2 is:

M_2=\dfrac{1}{11}\sum_{i=1}^{11}((-1)+12+(-1)+(-3)+3+(-5)+5+2+(-11)+(-1)+(-3))\\\\\\ M_2=\dfrac{-3}{11}=-0.273

s_2=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{11}(x_i-M)^2}\\\\\\

s_2=\sqrt{\dfrac{1}{10}\cdot [(-1-(-0.273))^2+(12-(-0.273))^2+...+(-3-(-0.273))^2]}\\\\\\

s_2=\sqrt{\dfrac{1}{10}\cdot [(0.53)+(150.62)+...+(115.07)+(0.53)+(7.44)]}\\\\\\s_2=\sqrt{\dfrac{348.182}{10}}=\sqrt{34.82}\\\\\\s_2=5.901

a) We have to calculate a 95% confidence interval for the difference between means, with a T-model.

The Group 1, of size n1=10 has a mean of 5 and a standard deviation of 8.743.

The Group 2, of size n2=11 has a mean of -0.273 and a standard deviation of 5.901.

The difference between sample means is Md=5.273.

M_d=M_1-M_2=5-(-0.273)=5.273

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{8.743^2}{10}+\dfrac{5.901^2}{11}}\\\\\\s_{M_d}=\sqrt{7.644+3.166}=\sqrt{10.81}=3.288

The degrees of freedom for this test are:

df=n_1+n_2-1=10+11-2=19

The critical t-value for a 95% confidence interval and 19 degrees of freedom is t=1.96.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_{M_d}=1.96 \cdot 3.288=6.444

Then, the lower and upper bounds of the confidence interval are:

LL=M_d-t \cdot s_{M_d} = 5.273-6.444=-1.171\\\\UL=M_d+t \cdot s_{M_d} = 5.273+6.444=11.717

The 95% confidence interval for the mean is (-1.171, 11.717).

b) This is a hypothesis test for the difference between populations means.

The claim is that increasing the amount of calcium in our diet reduce blood pressure.

Then, the null and alternative hypothesis are:

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

The significance level is 0.05.

The sample 1, of size n1=10 has a mean of 5 and a standard deviation of 8.743.

The sample 1, of size n1=11 has a mean of -0.273 and a standard deviation of 5.901.

The difference between sample means is Md=5.273.

M_d=M_1-M_2=5-(-0.273)=5.273

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{8.743^2}{10}+\dfrac{5.901^2}{11}}\\\\\\s_{M_d}=\sqrt{7.644+3.166}=\sqrt{10.81}=3.288

Then, we can calculate the t-statistic as:

t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{5.273-0}{3.288}=\dfrac{5.273}{3.288}=1.604

The degrees of freedom for this test are:

df=n_1+n_2-1=10+11-2=19

This test is a right-tailed test, with 19 degrees of freedom and t=1.604, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>1.604)=0.063

As the P-value (0.063) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that increasing the amount of calcium in our diet reduce blood pressure.

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Answer:

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Step-by-step explanation:

Given that the 2 triangles are congruent based on the Hypotenuse-leg theorem, this implies that:

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