A student hands each of 300 classmates 2 chocolates side by side on a plate. Of the 300 students, 171 choose the chocolate that¢s on their right hand side, and the remaining 129 choose the chocolate that¢s on their left. The student says, “That¢s just like tossing a coin.” The student¢s friend says, “No, it¢s not.” Help them settle their argument by performing a one-sample z test in the next Problems.

1) Find the P-value of the test.

To find the P-value of the test, we can use the one-sample z test formula.

First, we need to calculate the sample proportion for choosing the chocolate on the right-hand side. We divide the number of students who chose right-hand side chocolate (171) by the total number of students (300) to get the sample proportion:

p̂ = 171/300 = 0.57

Next, we need to calculate the expected proportion under the assumption that the choice of chocolates is random like flipping a coin. Since there are two chocolates and each has an equal chance of being chosen, the expected proportion is 0.5.

Now we can calculate the test statistic z:
z = (p̂ - p₀) / sqrt(p₀ * (1 - p₀) / n)

where p₀ is the expected proportion (0.5), and n is the sample size (300).

Plugging in the values, we get:
z = (0.57 - 0.5) / sqrt(0.5 * (1 - 0.5) / 300)
z = 1.702

Next, we need to find the p-value associated with the test statistic. This can be done by looking up the z-value in a standard normal distribution table or by using statistical software or a calculator.

Assuming a two-tailed test, the p-value is the probability of observing a z-score more extreme than the calculated z-score of 1.702. This means we need to find the probability of observing a z-score less than -1.702 or greater than 1.702.

Looking up the z-score in a standard normal distribution table or using a calculator, we find that the p-value is approximately 0.088.

Therefore, the P-value of the one-sample z test is 0.088.