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3 years ago

Part A: Consider the equation x + 7 = 16. Which number from the set {5, 7, 9, 11} makes the equation true?

Mathematics

1 answer:

wel3 years ago

3 0

Part A: Substitute/plug in each number into the equation to see which number will make the equation true.

x + 7 = 16 Plug in 5 into "x"

5 + 7 = 16

12 = 16 5 doesn't make the equation true because it equals 12 not 16

x + 7 = 16 Plug in 7 into "x"

7 + 7 = 16

14 = 16 7 doesn't make the equation true because it equals 14 not 16

x + 7 = 16 Plug in 9 into "x"

9 + 7 = 16

16 = 16 9 makes the equation true because it equals 16

Part B: Plug in 9 into "x" in the inequality to see if it still makes it true.

x + 7 < 16 [x plus 7 is less than 16]

9 + 7 < 16

16 < 16 [16 is less than 16] The same number would not make the inequality true because 16 can't be less than itself.

To figure out which numbers satisfy the inequality, plug it into the inequality:

x + 7 < 16 Plug in 5 into "x"

5 + 7 < 16

12 < 16 5 does satisfy the inequality because 12 is less than 16

x + 7 < 16 Plug in 7 into "x"

14 < 16 7 does satisfy the inequality because 14 is less than 16

x + 7 < 16 Plug in 11 into "x"

11 + 7 < 16

18 < 16 11 doesn't satisfy the inequality because 18 isn't less than 16

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For this case the statistic calculated is $t= 1.37$

$p_v =0.17$

So the p value is a very high value and using the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and there is enough evidence to conclude that we have a difference between the two means at 10% of significance.

Step-by-step explanation:

Data given and notation

$\bar X_{A}$ represent the mean for the sample for A

$\bar X_{B}$ represent the mean for the sample B

$s_{A}$ represent the sample standard deviation for the sample A

$s_{B}=$ represent the sample standard deviation for the sample B

$n_{A}=20$ sample size for the group poisoned

$n_{B}=20$ sample size for the group unpoisoned

t would represent the statistic (variable of interest)

Concepts and formulas to use

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

Null hypothesis: $\mu_{A} - \mu_{B} =0$

Alternative hypothesis: $\mu_{A} - \mu_{B} \neq 0$

For this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{A}-\bar X_{B}}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}}$ (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.

Calculate the statistic

For this case the statistic calculated is $t= 1.37$

Statistical decision

The significance level is 0.1 $\alpha=0.1$ , but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:

$df=n_{A}+n_{B}-2$

The p value on this case is given

$p_v =0.17$

So the p value is a very high value and using the significance level given $\alpha=0,1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and there is enough evidence to conclude that we have a difference between the two means at 10% of significance.

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