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Let n1equals50, Upper X 1equals30, n2equals50, and Upper X 2equals10. Complete parts (a) and (b) below. a. At the 0.05 leve

Viefleur [7K]

Answer:

Null hypothesis: $p_{1} = p_{2}$

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

$z=\frac{0.6-0.2}{\sqrt{0.4(1-0.4)(\frac{1}{50}+\frac{1}{50})}}=4.082$

$p_v =2*P(Z>4.082)=4.46x10^{-5}$

So the p value is a very low value and using any significance level for example $\alpha=0.05$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion 1 is significantly different from proportion 2.

Step-by-step explanation:

1) Data given and notation

$X_{1}=30$ represent the number of people with a characteristic in 1

$X_{2}=10$ represent the number of people with a characteristic in 2

$n_{1}=50$ sample of 1 selected

$n_{2}=50$ sample of 2 selected

$p_{1}=\frac{30}{50}=0.6$ represent the proportion of people with a characteristic in 1

$p_{2}=\frac{10}{50}=0.2$ represent the proportion of people with a characteristic in 2

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the proportion 1 is different from proportion 2 , 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 $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{30+10}{50+50}=0.4$

3) Calculate the statistic

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

$z=\frac{0.6-0.2}{\sqrt{0.4(1-0.4)(\frac{1}{50}+\frac{1}{50})}}=4.082$

4) 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 two sided test the p value would be:

$p_v =2*P(Z>4.082)=4.46x10^{-5}$

So the p value is a very low value and using any significance level for example $\alpha=0.05$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion 1 is significantly different from proportion 2.

3 0

3 years ago

Find the measurement

xenn [34]

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$6z + 7 = 8z - 9$