91Ó°ÊÓ

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$H_0:pâ‰\yen 0.6$

The alternative hypothesis is$H_a:\mathrm{μ}<0.6$

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 $\mathrm{Î\pm }=5\%=0.05$

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

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

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 $100(1-\mathrm{Î\pm })\%$confidence interval on the proportion p of the population that belongs to this class is

$p^{\text{'}}-z_{\frac{\mathrm{Î\pm }}{2}}\sqrt{\frac{p^{\text{'}}\left(1-p^{\text{'}}\right)}{n}}≤p≤p^{\text{'}}+z_{\frac{\mathrm{Î\pm }}{2}}\sqrt{\frac{p^{\text{'}}\left(1-p^{\text{'}}\right)}{n}}$

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

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

$\frac{\mathrm{Î\pm }}{2}=0.025⇒z_{\frac{\mathrm{Î\pm }}{2}}=z_{0.025}=1.96$

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

$0.53-1.96\sqrt{\sqrt{\frac{0.53(1-0.53)}{64}}}≤\mathrm{μ}≤0.53+1.96\sqrt{\frac{0.53(1-0.53)}{64}}$

$0.53-1.96×0.06≤\mathrm{μ}≤0.53+1.96×0.06$

Therefore, a CI on the population proportion is

$0.408≤\mathrm{μ}≤0.654$