Step 1 :Calculate the test statistic, z0, using the formula: \(z0 = \frac{{\hat{p} - p0}}{{\sqrt{{p0 * (1 - p0) / n}}}}\)
Step 2 :Substitute the given values into the formula: \(z0 = \frac{{0.35 - 0.3}}{{\sqrt{{0.3 * (1 - 0.3) / 100}}}}\)
Step 3 :Simplify the equation to get: \(z0 = \frac{{0.05}}{{\sqrt{{0.21 / 100}}}}\)
Step 4 :Further simplify to get: \(z0 = \frac{{0.05}}{{0.0458}}\)
Step 5 :Round the result to two decimal places to get: \(z0 = 1.09\)
Step 6 :Find the P-value, which is the probability of observing a test statistic as extreme as z0, given that the null hypothesis is true. Since this is a one-tailed test (H1: p > 0.3), find the area to the right of z0 = 1.09 on the standard normal distribution.
Step 7 :Using a standard normal (Z) table or a calculator, find that the area to the left of z0 = 1.09 is approximately 0.8621. Therefore, the area to the right (which is the P-value) is 1 - 0.8621 = 0.1379 (rounded to four decimal places).
Step 8 :Compare the P-value to the significance level, \(\alpha = 0.05\). If the P-value is less than \(\alpha\), reject the null hypothesis. If the P-value is greater than \(\alpha\), do not reject the null hypothesis.
Step 9 :Since 0.1379 > 0.05, do not reject the null hypothesis. Therefore, the final answer is: \(\boxed{\text{Do not reject the null hypothesis, because the P-value is greater than } \alpha}\)