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vampirchik [111]

3 years ago

9

Function 1: y = 4x + 5 Function 2: The line passing through the points (1, 6) and (3, 10). Which of these functions has the grea

ter rate of change? A) Function 1, because the slope is 5 and the slope of function 2 is 4. B) Function 1, because the slope is 4 and the slope of function 2 is 2. C) Function 2, because the slope is 7 and the slope of function 1 is 5. D) Function 2, because the slope is 5 and the slope of function 1 is 4.

1 answer:

Leona [35]3 years ago

7 0

Function 1 : y = 4x + 5....slope is 4

function 2 : (1,6)(3,10)
slope = (10 - 6) / (3 - 1) = 4/2 = 2....slope is 2

B. function 1, because the slope is 4 and the slope of function 2 is 2.

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Answer:

$z=\frac{0.659 -0.5}{\sqrt{\frac{0.5(1-0.5)}{502}}}=7.124$

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So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is significantly higher than 0.5.

Step-by-step explanation:

Data given and notation

n=502 represent the random sample taken

X=331 represent the adults that said they would not be bothered if the NAtional security agency

$\hat p=\frac{331}{502}=0.659$ estimated proportion of people who would not be bothered if the NAtional security agency

$p_o=0.5$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than the majority of 0.5:

Null hypothesis: $p\leq 0.5$

Alternative hypothesis: $p > 0.5$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.659 -0.5}{\sqrt{\frac{0.5(1-0.5)}{502}}}=7.124$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.01$ . The next step would be calculate the p value for this test.

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

$p_v =P(z>7.124)=5.24x10^{-13}$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is significantly higher than 0.5.

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