Question

In July of 1997, Australians were asked if they thought unemployment would increase, and 47% thought that it would increase. In November of 1997, they were asked again. At that time 284 out of 631 said that they thought unemployment would increase ("Morgan gallup poll," 2013). At the 5% level, is there enough evidence to show that the proportion of Australians in November 1997 who believe unemployment would increase is less than the proportion who felt it would increase in July 1997?

Answer 1

`z=\frac{0.45 -0.47}{\sqrt{(0.47(1-0.47))/(631)}}=-1.007`

Now we can find the p value with the alternative hypothesis and using this probability:

`p_v =P(z<-1.007)=0.157`

Since the p value is higher than the significance level given of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true proportion of interest is not significantly lower than 0.47

Explanation:

Information given

n=631 represent the random sample selected

X=284 represent the people who said that they thought unemployment would increase

`\hat p=(284)/(631)=0.45` estimated proportion of people who said that they thought unemployment would increase

`p_o=0.47` is the value that we want to test

`\alpha=0.05` represent the significance level

z would represent the statistic

Alternative hypothesis:
`p < 0.47`

The statistic would be given by:

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

Replacing the info given we got:

`z=\frac{0.45 -0.47}{\sqrt{(0.47(1-0.47))/(631)}}=-1.007`

Now we can find the p value with the alternative hypothesis and using this probability:

`p_v =P(z<-1.007)=0.157`

Since the p value is higher than the significance level given of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true proportion of interest is not significantly lower than 0.47