Answer :
To test whether the proportion of puts and shorts among all trades has increased, we can perform a hypothesis test for the difference in proportions.
Let's define the following:
\( p_1 \) = Proportion of puts and shorts in the past 5 years.
\( p_2 \) = Proportion of puts and shorts among today's trades.
We want to test the null hypothesis (\( H_0 \)) that there is no difference in the proportions of puts and shorts in the past 5 years compared to today's trades. The alternative hypothesis (\( H_1 \)) is that the proportion of puts and shorts among today's trades is greater than in the past 5 years.
\[ H_0: p_2 \leq p_1 \]
\[ H_1: p_2 > p_1 \]
We'll conduct a one-tailed test because we want to determine if the proportion of puts and shorts among today's trades is significantly greater than in the past 5 years.
Now, let's calculate the sample proportions:
- In the past 5 years: \( p_1 = 0.63 \) (given)
- Today's trades: \( p_2 = \frac{435}{631} \)
First, let's calculate \( p_2 \):
\[ p_2 = \frac{435}{631} \approx 0.689 \]
Now, let's calculate the test statistic:
\[ Z = \frac{p_2 - p_1}{\sqrt{\frac{p_1(1-p_1)}{n}}} \]
Where \( n \) is the sample size.
Given:
- \( p_1 = 0.63 \)
- \( p_2 = 0.689 \)
- \( n = 631 \)
\[ Z = \frac{0.689 - 0.63}{\sqrt{\frac{0.63(1-0.63)}{631}}} \]
Now, we calculate the value of \( Z \):
\[ Z = \frac{0.689 - 0.63}{\sqrt{\frac{0.63(0.37)}{631}}} \]
\[ Z = \frac{0.059}{\sqrt{\frac{0.63(0.37)}{631}}} \]
\[ Z \approx \frac{0.059}{\sqrt{\frac{0.2321}{631}}} \]
\[ Z \approx \frac{0.059}{\sqrt{0.000368}} \]
\[ Z \approx \frac{0.059}{0.0192} \]
\[ Z \approx 3.068 \]
Now, let's find the critical value for a one-tailed test at 5% significance level. Since this is an upper-tailed test, we're interested in finding the critical value that corresponds to 5% in the upper tail of the standard normal distribution.
Using a standard normal distribution table or a calculator, we find that the critical value for a one-tailed test at 5% significance level is approximately 1.645.
Now, let's find the p-value corresponding to the test statistic \( Z = 3.068 \).
Using a standard normal distribution table or a calculator, the p-value is very close to 0.001.
Now, let's state the conclusion:
Since the p-value (approximately 0.001) is less than the significance level of 0.05, we reject the null hypothesis. Therefore, at 5% significance level, we have sufficient evidence to conclude that the proportion of puts and shorts among all trades has increased.
