Mathematics, 04.03.2020 02:25 melekatoa

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 .

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Answer from: mbrisen3396

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

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 $p_v < \alpha$

Answer from: Quest

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like, however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

$\bf e(\stackrel{x_1}{-2}~,~\stackrel{y_1}{2})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{0-2}{)}\implies \implies \cfrac{-2}{0+2}\implies \cfrac{-2}{2}\implies -1 \\\\[-0.35em] ~\dotfill$

$\bf d(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{)}{)}\implies \cfrac{0+1}{0+2}\implies \cfrac{1}{2} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope~of~ef}{-1\implies \cfrac{-1}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{-1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{1}{1}\implies 1}}$