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Lera25 [3.4K]
2 years ago
11

I just need help with the last part. the c) part Describe the end behavior of the function:

Mathematics
2 answers:
Mkey [24]2 years ago
8 0
The end has a more entisified curve than the beginning
IRINA_888 [86]2 years ago
4 0
I'm not sure but I think that you would describe it as positive or steep.
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Step-by-step explanation:

Because halfway of 4, -2 is 0

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b = 13

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A survey was conducted in the United Kingdom, where respondents were asked if they had a university degree. One question asked,
katovenus [111]

Answer:

z=\frac{0.121-0.0892}{\sqrt{0.101(1-0.101)(\frac{1}{373}+\frac{1}{639})}}=1.620  

p_v =2*P(Z>1.620)=0.105  

If we compare the p value and using any significance level for example \alpha=0.05 always p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the two proportions are not statistically different at 5% of significance

Step-by-step explanation:

Data given and notation  

X_{1}=45 represent the number of correct answers for university degree holders

X_{2}=57 represent the number of correct answers for university non-degree holders  

n_{1}=373 sample 1 selected

n_{2}=639 sample 2 selected

p_{1}=\frac{45}{373}=0.121 represent the proportion of correct answers for university degree holders  

p_{2}=\frac{57}{639}=0.0892 represent the proportion of correct answers for university non-degree holders  

z would represent the statistic (variable of interest)  

p_v represent the value for the test (variable of interest)

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the proportions are different between the two groups, the system of hypothesis would be:  

Null hypothesis:p_{1} = p_{2}  

Alternative hypothesis:p_{1} \neq p_{2}  

We need to apply a z test to compare proportions, and the statistic is given by:  

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{45+57}{373+639}=0.101

Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

z=\frac{0.121-0.0892}{\sqrt{0.101(1-0.101)(\frac{1}{373}+\frac{1}{639})}}=1.620  

Statistical decision

For this case we don't have a significance level provided \alpha, but we can calculate the p value for this test.  

Since is a one side test the p value would be:  

p_v =2*P(Z>1.620)=0.105  

If we compare the p value and using any significance level for example \alpha=0.05 always p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the two proportions are not statistically different at 5% of significance

7 0
2 years ago
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