Your statistics instructor claims that $60$percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey $64$of her past Elementary Statistics students and find that $34$feel more enriched as a result of her class. Now, what do you think?

The null hypothesis states that there is no difference between a proportion and the proportion of the population.

From the question itself, we are given$60\%$of the students took Elementary Statistics class for feeling more enriched. So, the null hypothesis is$H_0:p\ge 0.6$

The alternative hypothesis is$H_a:\mu <0.6$

The random variable $P$is the proportion of students who feel more enriched as a result of taking Elementary Statistics.

The distribution to use for the test is a Normal distribution with parameters $p=0.6$

Therefore, the distribution is

$\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.6(1-0.6)}{64}}$

The test statistic for this test is

$$\begin{gathered} z=\frac{\left|p^{\text{'}}-p\right|}{\sqrt{p(1-p)/n}} \\ z=\frac{|0.53-0.6|}{\sqrt{0.24/64}} \\ z=1.12 \end{gathered}$$

where,

$$\begin{gathered} p^{\text{'}}=\frac{x}{n} \\ p=\frac{34}{64} \\ p=0.53. \end{gathered}$$

So, the p- value for this test is

$$\begin{gathered} \mathrm{p}-\mathrm{value}=P\left\{p^{\text{'}}<0.53\right\} \\ =0.1308 \end{gathered}$$

Level of confidence $\alpha =5\%=0.05$

Decision: We fail to reject reject $H_0$

Reason for decision: When the p-value is greater than the established alpha value we do not reject the null hypothesis. Now, we see that

$\mathrm{p}\text{-value}=0.1308>0.05=\alpha .$

Conclusion: At a level of significance of $5\%$, there is sufficient evidence to conclude that less than $60\%$of her students feel more enriched.

If p is the proportion of observations in a random sample of size n that belongs to a class of interest, an approximate $100(1-\alpha )\%$confidence interval on the proportion p of the population that belongs to this class is

$p^{\text{'}}-z_{\frac{\alpha }{2}}\sqrt{\frac{p^{\text{'}}\left(1-p^{\text{'}}\right)}{n}}\le p\le p^{\text{'}}+z_{\frac{\alpha }{2}}\sqrt{\frac{p^{\text{'}}\left(1-p^{\text{'}}\right)}{n}}$

where,role="math" localid="1650374329053" $z_{\frac{a}{2}}$is the upper limit and $\frac{\alpha }{2}$percentage point of the standard normal distribution.

We need to find $95\%$CI on the population proportion, then

$\frac{\alpha }{2}=0.025\Rightarrow z_{\frac{\alpha }{2}}=z_{0.025}=1.96$

From equation (i) and (ii) we get,

$0.53-1.96\sqrt{\sqrt{\frac{0.53(1-0.53)}{64}}}\le \mu \le 0.53+1.96\sqrt{\frac{0.53(1-0.53)}{64}}$

$0.53-1.96\times 0.06\le \mu \le 0.53+1.96\times 0.06$

Therefore, a CI on the population proportion is

$0.408\le \mu \le 0.654$