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3 years ago

Choose the true statement. 0-2 <-3 0-2>-3 0-2 = -3

Mathematics

1 answer:

Natasha2012 [34]3 years ago

3 0

The true statement is -2>-3

Step-by-step explanation:

The given statement included inequality and comparison symbols. The numbers involved are negative integers.

As the number increases its value decreases which means that -1 will be greater than -2 and so on

Given statement will be discussed one by one

-2 <-3

-2 comes before -3 so the statement is true as -2 will be greater than -3.

-2>-3

The statement is true because -2 is greater than -3

The true statement is -2>-3

Keywords: Comparisons, Inequality

Learn more about inequalities at:

#LearnwithBrainly

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$t=\frac{85.8-113.4}{\sqrt{\frac{(50)^2}{145}+\frac{(75)^2}{141}}}}=-3.65$

$p_v =2*P(t_{(284)}$

Using Excel we can use the following code: "=2*T.DIST(-3.65;284;TRUE)"

So the p value is a very low value and using any significance level giveen $\alpha=0.05$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the means are different.

Step-by-step explanation:

1) Data given and notation

$\bar X_{M}=85.8$ represent the mean for the males

$\bar X_{F}=113.4$ represent the mean for females

$s_{M}=50$ represent the sample standard deviation for males

$s_{F}=75$ represent the sample standard deviation for female

$n_{M}=145$ sample size for the group poisoned

$n_{F}=141$ sample size for the group unpoisoned

t would represent the statistic (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for the two groups are equal or not , the system of hypothesis would be:

Null hypothesis: $\mu_{M} = \mu_{F}$

Alternative hypothesis: $\mu_{M} \neq \mu_{F}$

If we analyze the size for the samples both are higher than 30 but the population deviations are not given, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{M}-\bar X_{F}}{\sqrt{\frac{s^2_{M}}{n_{M}}+\frac{s^2_{F}}{n_{F}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

3) Calculate the statistic

We can replace in formula (1) the results obtained like this:

$t=\frac{85.8-113.4}{\sqrt{\frac{(50)^2}{145}+\frac{(75)^2}{141}}}}=-3.65$

4) Statistical decision

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

$df=n_{M}+n_{F}-2=145+141-2=284$

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

$p_v =2*P(t_{(284)}$

Using Excel we can use the following code: "=2*T.DIST(-3.65;284;TRUE)"

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