Question 1150934
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Let group 1 be the sample of Republicans. Group 2 will be the sample of Democrats.


H0: *[Tex \Large p_{1} = p_{2}] (equivalent to *[Tex p_{1}-p_{2} = 0]) is the null hypothesis
H1: *[Tex \Large p_{1} < p_{2}] (equivalent to *[Tex p_{1}-p_{2} < 0]) is the alternative hypothesis
where,
*[Tex \Large p_{1}] = population proportion of Republicans in favor of lowering the standards
*[Tex \Large p_{2}] = population proportion of Democrats in favor of lowering the standards


We are conducting a left tailed test because of the "less than" sign in the alternative hypothesis.  


We are given
*[Tex \Large x_{1} = 200]
*[Tex \Large x_{2} = 168]
as the two counts of successes (ie the number of people who responded in favor for each group). And also, we have these sample sizes
*[Tex \Large n_{1} = 1000]
*[Tex \Large n_{2} = 800]


Which helps us find the sample proportions for each group
*[Tex \Large \hat{p}_{1} = \frac{x_1}{n_1} = \frac{200}{1000} = 0.2] is the sample proportion of Republicans in favor of lowering the standards
*[Tex \Large \hat{p}_{2} = \frac{x_2}{n_2} = \frac{168}{800} = 0.21] is the sample proportion of Democrats in favor of lowering the standards
Based on these two p-hat values, it does look like there is a higher proportion of Democrats in favor of lowering the standards. However, let's continue with the hypothesis test.


Compute the pooled proportion. The bar over the 'p' indicates an average.
*[Tex \Large \overline{p} = \frac{x_1+x_2}{n_1+n_2} = \frac{200+168}{1000+800} \approx 0.204444444] 


Now onto the Standard Error. 
*[Tex \Large SE = \sqrt{\overline{p}*(1-\overline{p})*\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}]
*[Tex \Large SE \approx \sqrt{0.204 444 444*(1-0.204444444)*\left(\frac{1}{1000}+\frac{1}{800}\right)}]
*[Tex \Large SE \approx 0.019129965] is the approximate standard error


Compute the test statistic
*[Tex \Large z = \frac{ (\hat{p}_{1}-\hat{p}_{2}) - (p_{1}-p_{2}) }{SE}]


*[Tex \Large z \approx \frac{ (0.2 - 0.21) - (0) }{0.019129965}]


*[Tex \Large z \approx -0.522740110]


*[Tex \Large z \approx -0.52]


Use a Z table similar to this one
<a href = "http://www.z-table.com/">http://www.z-table.com/</a>
To find that *[Tex \Large P(Z < -0.52) \approx 0.3015]


The P-value is approximately 0.3015


This is much larger than alpha = 0.02, so we fail to reject the null hypothesis.
We do not have enough statistically significant evidence to overturn or reject the null. 
Therefore, we must accept that the population proportions p1 and p2 are the same. 
In other words, the proportion of people in favor of lowering the standards are the same for each group. 


Going back to the question "<font color=blue>can we conclude that there is a larger proportion of Democrats in favor of lowering the standards?</font>", the answer is "no, we cannot conclude there is a larger proportion of Democrats in favor of lowering the standards".
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