Calculate the sample proportion $\hat{p}$:

$$\hat{p}=\cfrac{X}{n}=\cfrac{46}{96}=0.4792$$
When using $\pi_0$ to calculate the margin of error, the general formula for a $C\%\,CI$ for the population proportion $\pi$ is:
$$CI_{\pi}=\bigg(\hat{p}- z^*\cdot \sqrt{\cfrac{\pi_0\cdot(1-\pi_0)}{n}},\,\,\,\, \hat{p} + z^*\cdot \sqrt{\cfrac{\pi_0\cdot(1-\pi_0)}{n}} \bigg)$$
For a given confidence level $C$ (in $\%$), the critical value $z^*$ of the standard normal distribution is the value such that $\mathbb{P}(-z^* \leq Z \leq z^*)=\cfrac{C}{100}$.



To calculate this critical value $z^*$ in Excel, make use of the following function:

NORM.INV(probability, mean, standard_dev)

• probability: A probability corresponding to the normal distribution.
• mean: The mean of the distribution.
• standard_dev: The standard deviation of the distribution.

Here, we have $C=94$. Thus, to calculate $z^*$ such that $\mathbb{P}(-z^* \leq Z \leq z^*)=0.94$, run the following command:

$$\begin{array}{c} =\text{NORM.INV}((100+C)/200, 0, 1)\\ \downarrow\\ =\text{NORM.INV}(194/200, 0, 1) \end{array}$$
This gives:
$$z^* = 1.8808$$
When using $\pi_0$ to calculate the margin of error, we calculate the lower bound $L$ of the confidence interval:
$$L = \hat{p} - z^*\cdot \sqrt{\cfrac{\pi_0\cdot(1-\pi_0)}{n}} = 0.4792 - 1.8808 \cdot \sqrt{\cfrac{0.55 \cdot (1-0.55)}{96}} = 0.384$$
When using $\pi_0$ to calculate the margin of error, we calculate the lower bound $U$ of the confidence interval:
$$U = \hat{p} + z^*\cdot \sqrt{\cfrac{\pi_0\cdot(1-\pi_0)}{n}} = 0.4792 + 1.8808 \cdot \sqrt{\cfrac{0.55 \cdot (1-0.55)}{96}} = 0.575$$
Thus, the $94\%$ confidence interval for the population proportion $\pi$ is:
$$CI_{\pi,\,94\%}=(0.384,\,\,\, 0.575)$$
Based on this confidence interval, the null hypothesis $H_0: \pi = 0.55$ should not be rejected at the $\alpha = 0.06$ level of significance because $\pi_0 = 0.55$ falls inside the confidence interval.

Calculate the sample proportion $\hat{p}$:

$$\hat{p}=\cfrac{X}{n}=\cfrac{46}{96}=0.4792$$
When using $\pi_0$ to calculate the margin of error, the general formula for a $C\%\,CI$ for the population proportion $\pi$ is:
$$CI_{\pi}=\bigg(\hat{p}- z^*\cdot \sqrt{\cfrac{\pi_0\cdot(1-\pi_0)}{n}},\,\,\,\, \hat{p} + z^*\cdot \sqrt{\cfrac{\pi_0\cdot(1-\pi_0)}{n}} \bigg)$$
For a given confidence level $C$ (in $\%$), the critical value $z^*$ of the standard normal distribution is the value such that $\mathbb{P}(-z^* \leq Z \leq z^*)=\cfrac{C}{100}$.


To calculate this critical value $z^*$ in R, make use of the following function:

qnorm(p, mean, sd, lower.tail)

• p: A probability corresponding to the normal distribution.
• mean: The mean of the distribution.
• sd: The standard deviation of the distribution.
• lower.tail: If TRUE (default), probabilities are $\mathbb{P}(X \leq x)$, otherwise, $\mathbb{P}(X \gt x)$.

Here, we have $C=94$. Thus, to calculate $z^*$such that $\mathbb{P}(-z^* \leq Z \leq z^*)=0.94$, run the following command:

$$\begin{array}{c} \text{qnorm}(p = (100+C)/200, mean = 0, sd = 1, lower.tail = \text{TRUE})\\ \downarrow\\ \text{qnorm}(p =194/200, mean = 0, sd = 1, lower.tail = \text{TRUE}) \end{array}$$
This gives:
$$z^* = 1.8808$$
When using $\pi_0$ to calculate the margin of error, we calculate the lower bound $L$ of the confidence interval:
$$L = \hat{p} - z^*\cdot \sqrt{\cfrac{\pi_0\cdot(1-\pi_0)}{n}} = 0.4792 - 1.8808 \cdot \sqrt{\cfrac{0.55 \cdot (1-0.55)}{96}} = 0.384$$
When using $\pi_0$ to calculate the margin of error, we calculate the lower bound $U$ of the confidence interval:
$$U = \hat{p} + z^*\cdot \sqrt{\cfrac{\pi_0\cdot(1-\pi_0)}{n}} = 0.4792 + 1.8808 \cdot \sqrt{\cfrac{0.55 \cdot (1-0.55)}{96}} = 0.575$$
Thus, the $94\%$ confidence interval for the population proportion $\pi$ is:
$$CI_{\pi,\,94\%}=(0.384,\,\,\, 0.575)$$
Based on this confidence interval, the null hypothesis $H_0: \pi = 0.55$ should not be rejected at the $\alpha = 0.06$ level of significance because $\pi_0 = 0.55$ falls inside the confidence interval.