Step 1 :State the null hypothesis H0: \(p = 0.20\) and the alternative hypothesis Ha: \(p > 0.20\)
Step 2 :Plan to use the one-sample z-test for a proportion
Step 3 :Calculate the pooled sample proportion (p) and the standard error (SE) using the formulae \(p = \frac{51}{233} = 0.219\) and \(SE = \sqrt{\frac{p * (1 - p)}{n}} = \sqrt{\frac{0.219 * (1 - 0.219)}{233}} = 0.028\)
Step 4 :Compute the z-score test statistic (z) using the formula \(z = \frac{p - P}{SE} = \frac{0.219 - 0.20}{0.028} = 0.68\)
Step 5 :Find the P-value associated with the observed z-score (0.68) in the Normal Distribution, which is 0.248
Step 6 :\(\boxed{\text{Since the P-value (0.248) is greater than the significance level (0.05), we cannot reject the null hypothesis. Therefore, there is not sufficient evidence to warrant support of the claim that more than 20% of users develop nausea.}}\)