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"A random sample of 125 registered voters in Phoenix is asked if they favor the use of Oxygenated fuels year-round to reduce air pollution and 86 voters responded positively. The city office claims that more than 65% of voters in Phoenix use oxygenated fuels. At 5% level of significance, to test that more than 65% of voters in Phoenix used oxygenated fuels, the suitable null and alternative hypotheses are"

User Mike Yan
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1 Answer

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Answer:


z=\frac{0.688 -0.65}{\sqrt{(0.65(1-0.65))/(125)}}=0.891


p_v =P(z>0.891)=0.1865

The p value for this case is higher than the significance level provided
\alpha=0.05 so we don't have enough evidence to reject the null hypothesis and we can't conclude that the true proportion is higher than 0.65

Explanation:

Data given and notation

n=125 represent the random sample taken

X=86 represent the voters who respond positively


\hat p=(86)/(125)=0.688 estimated proportion of voters who respond positively


p_o=0.65 is the value that we want to test


\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)


p_v represent the p value (variable of interest)

System of hypothesis

We need to conduct a hypothesis in order to test the claim that true proportion is higher than 0.65, and the system of hypothesis are:

Null hypothesis:
p\leq 0.65

Alternative hypothesis:
p > 0.65

The statistic is given by:


z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}} (1)

Replacing we got:


z=\frac{0.688 -0.65}{\sqrt{(0.65(1-0.65))/(125)}}=0.891

Conclusion

The significance level provided
\alpha=0.05.

Since is a right tailed test the p value would be:


p_v =P(z>0.891)=0.1865

The p value for this case is higher than the significance level provided
\alpha=0.05 so we don't have enough evidence to reject the null hypothesis and we can't conclude that the true proportion is higher than 0.65

User NcAdams
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