Solve quadratic equation (x-3)^2+5=14

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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