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2 years ago

Evaluate |-11| + |-7|. -18 -4 4 18

Mathematics

1 answer:

Marta_Voda [28]2 years ago

4 0

Answer: D) 18

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Explanation:

|-11| simplifies to 11

|-7| simplifies to 7

Whatever is inside the absolute value, you just remove the negative and that's the result of that absolute value expression. It represents the distance on a number line. So for instance -11 is 11 units from 0, which is why |-11| = 11.

Overall,

|-11| + |-7| = 11 + 7 = 18

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Answer:

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

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

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If we use a significance level of 0.05 we see that $p_v > \alpha$ and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

Step-by-step explanation:

Data given and notation

$\bar X=5.63$ represent the sample mean

$s=1.61$ represent the standard deviation for the sample

$n=15$ sample size

$\mu_o =15/tex] represent the value that we want to test 
[tex]\alpha$ represent the significance level for the hypothesis 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 to be tested

We need to conduct a hypothesis in order to determine if the average is more than 5.63, the system of hypothesis would be:

Null hypothesis: $\mu \leq 5.63$

Alternative hypothesis: $\mu > 5.63$

Compute the test statistic

We don't know the population deviation, so for this case 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".

We can replace in formula (1) the info given like this:

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

Now we need to find the degrees of freedom for the t distribution given by:

$df=n-1=15-1=14$

P value

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

$p_v =P(t_{14}>1.516)=0.076$

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