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qwelly [4]
3 years ago
11

he one‑sample t statistic from a sample of n = 23 observations for the two‑sided test of H 0 : μ = 15 versus H α : μ > 15 has

the value t = 2.24 . Based on this information: 0.01 < P ‑ value < 0.025 and we would reject the null hypothesis at α = 0.025 are both correct. P ‑ value > 0.1 . we would reject the null hypothesis at α = 0.025 . 0.01 < P ‑ value < 0.025 .
Mathematics
1 answer:
DedPeter [7]3 years ago
5 0

Answer:

t = 2.24

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

df=n-1=23-1=22  

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

p_v =P(t_{(22)}>2.24)=0.01776  

And for this case we can conclude that:

0.01 < p_v < 0.025

And we will reject the null hypothesis at \alpha=0.025 since p_v < \alpha

Step-by-step explanation:

Data given and notation  

\bar X represent the mean height for the sample  

s represent the sample standard deviation

n=23 sample size  

\mu_o =15 represent the value that we want to test

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 15, the system of hypothesis would be:  

Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

If we analyze the size for the sample is > 30 but 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

For this case the statistic is given:

t = 2.24

P-value

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

df=n-1=23-1=22  

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

p_v =P(t_{(22)}>2.24)=0.01776  

And for this case we can conclude that:

0.01 < p_v < 0.025

And we will reject the null hypothesis at \alpha=0.025 since p_v < \alpha

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P(x) = x + 1x² – 34x + 343<br> d(x)= x + 9
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Answer:

x=\frac{9}{d-1},\:P=\frac{-297d+378}{\left(d-1\right)^2}+343

Step-by-step explanation:

Let us start by isolating x for dx = x + 9.

dx - x = x + 9 - x > dx - x = 9.

Factor out the common term of x > x(d - 1) = 9.

Now divide both sides by d - 1 > \frac{x\left(d-1\right)}{d-1}=\frac{9}{d-1};\quad \:d\ne \:1. Go ahead and simplify.

x=\frac{9}{d-1};\quad \:d\ne \:1.

Now, \mathrm{For\:}P=x+1x^2-34x+343, \mathrm{Subsititute\:}x=\frac{9}{d-1}.

P=\frac{9}{d-1}+1\cdot \left(\frac{9}{d-1}\right)^2-34\cdot \frac{9}{d-1}+343.

Group the like terms... 1\cdot \left(\frac{9}{d-1}\right)^2+\frac{9}{d-1}-34\cdot \frac{9}{d-1}+343.

\mathrm{Add\:similar\:elements:}\:\frac{9}{d-1}-34\cdot \frac{9}{d-1}=-33\cdot \frac{9}{d-1} > 1\cdot \left(\frac{9}{d-1}\right)^2-33\cdot \frac{9}{d-1}+343.

Now for 1\cdot \left(\frac{9}{d-1}\right)^2 > \mathrm{Apply\:exponent\:rule}: \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c} > \frac{9^2}{\left(d-1\right)^2} = 1\cdot \frac{9^2}{\left(d-1\right)^2}.

\mathrm{Multiply:}\:1\cdot \frac{9^2}{\left(d-1\right)^2}=\frac{9^2}{\left(d-1\right)^2}.

Now for 33\cdot \frac{9}{d-1} > \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} > \frac{9\cdot \:33}{d-1} > \frac{297}{d-1}.

Thus we then get \frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}+343.

Now we want to combine fractions. \frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}.

\mathrm{Compute\:an\:expression\:comprised\:of\:factors\:that\:appear\:either\:in\:}\left(d-1\right)^2\mathrm{\:or\:}d-1 > This\: is \:the\:LCM > \left(d-1\right)^2

\mathrm{For}\:\frac{297}{d-1}:\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:d-1 > \frac{297}{d-1}=\frac{297\left(d-1\right)}{\left(d-1\right)\left(d-1\right)}=\frac{297\left(d-1\right)}{\left(d-1\right)^2}

\frac{9^2}{\left(d-1\right)^2}-\frac{297\left(d-1\right)}{\left(d-1\right)^2} > \mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}> \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}

\frac{9^2-297\left(d-1\right)}{\left(d-1\right)^2} > 9^2=81 > \frac{81-297\left(d-1\right)}{\left(d-1\right)^2}.

Expand 81-297\left(d-1\right) > -297\left(d-1\right) > \mathrm{Apply\:the\:distributive\:law}: \:a\left(b-c\right)=ab-ac.

-297d-\left(-297\right)\cdot \:1 > \mathrm{Apply\:minus-plus\:rules} > -\left(-a\right)=a > -297d+297\cdot \:1.

\mathrm{Multiply\:the\:numbers:}\:297\cdot \:1=297 > -297d+297 > 81-297d+297 > \mathrm{Add\:the\:numbers:}\:81+297=378 > -297d+378 > \frac{-297d+378}{\left(d-1\right)^2}

Therefore P=\frac{-297d+378}{\left(d-1\right)^2}+343.

Hope this helps!

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