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Gemiola [76]
3 years ago
8

Sofia wants to place a sticker 2 ½ inches long in the center of a switch box that is 3 ¾ inches wide. About how far from the edg

e will she place the sticker?
Mathematics
1 answer:
Andreas93 [3]3 years ago
5 0
3 3/4 -2 1/2=
15/4-5/2=
15/4-5/2*2/2=
15/4-10/4=
5/4=

5/4 divided by 2 (gives you the distance from the edge if you put the sticker in the center)

5/4 / 2/1=5/8 inches from the edge

I think this is correct. Hope it helps
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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,
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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

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