Use the results from a survey of a simple random sample of 1040 adults. Among the 1040 respondents, $92\mathrm{\%}$ rated themselves as above average drivers. We want to test the claim that more than $\frac{4}{5}$ of adults rate themselves as above average drivers. Complete parts (a) through (e).
A. The critical value method is not always equivalent to the other two.
B. The confidence interval method is not always equivalent to the other two.
C. The P-value method is not always equivalent to the other two.
D. All three methods are always equivalent to each other.
c. It was stated that we can easily remember how to interpret P-values with this: "If the P is low, the null must go." What does this mean?
This statement means that if the is low in relation to then one should
the

Step 1 ：We are given a simple random sample of 1040 adults, among which 92% rated themselves as above average drivers. We want to test the claim that more than $\frac{4}{5}$ of adults rate themselves as above average drivers.

Step 2 ：We perform a hypothesis test. The null hypothesis (H0) is that the proportion of adults who rate themselves as above average drivers is equal to $\frac{4}{5}$, and the alternative hypothesis (H1) is that the proportion is greater than $\frac{4}{5}$.

Step 3 ：We use the sample proportion ($\hat{p}$ = 0.92) and the sample size (n = 1040) to calculate the test statistic and the p-value.

Step 4 ：The test statistic (z) is calculated using the formula $z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$, where $p_0$ is the hypothesized population proportion under the null hypothesis. Substituting the given values, we get $z\approx 9.67$.

Step 5 ：The p-value is the probability of observing a sample proportion as extreme as 0.92 or more extreme, given that the null hypothesis is true. In this case, the p-value is approximately 0.

Step 6 ：If the p-value is less than the significance level (usually 0.05), we reject the null hypothesis. Here, since the p-value is 0, which is less than 0.05, we reject the null hypothesis.

Step 7 ：Therefore, we conclude that more than $\frac{4}{5}$ of adults rate themselves as above average drivers.

Step 8 ：The final answer is the test statistic and the p-value, which are approximately 9.67 and 0 respectively. So, the final answer is $\boxed{{\displaystyle 9.67,0}}$.