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2 years ago

13

Compare using < >, or = 5 3/4 ____ 6 1/4

2 answers:

gregori [183]2 years ago

8 0

Answer:

<

Step-by-step explanation:

uysha [10]2 years ago

5 0

I was going to say that you would need to change the mixed numbers into an improper fraction but then I realized that they both have the same denominator (4), meaning that in this case if the whole number outside of the fraction is bigger than the one for the other then it’s greater than regardless.

6 is greater than 5, so

5¾<6¼

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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

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