Help Please! Find Sample test statistic and P-value

Question

In a fishing lodge brochure, the lodge advertises that 75% of its guests catch northern pike over 20 pounds. Suppose that last summer 71 out of a random sample of 86 guests did, in fact, catch northern pike weighing over 20 pounds. Does this indicate that the population proportion of guests who catch pike over 20 pounds is different from 75% (either higher or lower)? Use 𝛼 = 0.05.

Answer 1

The first thing you should do is to define parameters and set up your null and alternative hypotheses. Let p denote the true population proportion of guests who catch pike over 20 pounds. Following the problem statement, the null hypothesis is p = 0.75 and the alternative hypothesis is p != 0.75.

Next, you can calculate a test statistic for proportions. Let p_hat denote the sample proportion. So, p_hat is equal to 71/86. The test statistic for proportions is as follows: (p_hat - p)/sqrt((p*(1-p))/n) (it may be easier to look up the formula). You can replace p with the value you set in the null hypothesis and n with the number of samples.

Once you have calculated your test statistic, you can look up a z-table or use statistical software to determine the p-value and decide whether or not to reject the null hypothesis.