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$x=16$

$n=60$

Determine the hypotheses:

$H_0:p=20\%$

$=0.20$

$H_a:p>0.20$

The sample proportion is the number of successes divided by the sample size:

localid="1650442739498" $$\begin{gathered} \hat{p}=\frac{x}{n} \\ =\frac{16}{60} \end{gathered}$$

$鈮�0.2667$

Determine the value of the test-statistic:

localid="1650442753640" $$\begin{gathered} z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}} \\ =\frac{0.2667-0.2}{\sqrt{\frac{0.2(1-0.2)}{60}}} \end{gathered}$$

$鈮�1.29$

The $P-$value is the chance of getting the test statistic's result, or a number that is more severe. Calculate the $P-$value in table A as follows:

$P=P(Z>1.29)$

$=P(Z<-1.29)$

$=0.0985$

Reject the null hypothesis if the$P-$value is less than the significance level:

$P>0.05鈬�\text{Fail to reject}{H}_0$

$H_0:p=20\%$

$=0.20$

$H_a:p>0.20$

Type I error: Reject the null hypothesis $H_0$, when $H_0$ is true.

Consequence: Less people are willing to pay $\$100$ than it appears from the results of the test.

Type II error: Fail to reject the null hypothesis $H_0$, when $H_0$ is false.

Consequence: More people are willing to pay $\$100$than it appears from the results of the test.

Type I error is worse, because then the upgrade is less likely to be profitable.

For the upgrade to be profitable, the company needs to sell it to more than $20\%$ of their customers.

You can increase the power by:

Increasing the sample size (because having more information about the population will allow us to make better estimations).

Increase the significance level (because this increases the probability of making a Type I error and decreases the probability of making a Type II error; Since the power is $1$ decreased by the probability of making a Type II error and thus the power increases).

Making the alternative proportion $p$ more extreme (thus $p$ greater than $0.45$, since more extreme alternatives are easier to prove).