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1 year ago

7

Four points are labeled on the number line. M K L zo 0.5 1 Which point best represents 3? F. Point K G. H. Point 2 Point M Point

N J.

1 answer:

viva [34]1 year ago

8 0

The point that best represents 1/3 is point M .

The number line ranges from 0 to 0.5 with 10 divi

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zmey [24]

Answer:

$t=\frac{7.03-3.58}{\frac{9.42}{\sqrt{20}}}=1.638$

$p_v =P(t_{(19)}>1.638)=0.0589$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true mean is not significantly higher than 3.58 at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=7.03$ represent the sample mean

$s=9.42$ represent the sample standard deviation

$n=20$ sample size

$\mu_o =3.58$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

z 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 3.58 :

Null hypothesis: $\mu \leq 3.58$

Alternative hypothesis: $\mu > 3.58$

Since we don't know the population deviation, 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 can replace in formula (1) the info given like this:

$t=\frac{7.03-3.58}{\frac{9.42}{\sqrt{20}}}=1.638$

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=20-1=19$

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

$p_v =P(t_{(19)}>1.638)=0.0589$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true mean is not significantly higher than 3.58 at 5% of signficance.

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