Question

"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"

Answer 1

`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