5

1 point

Over the past 5 years, 63% of asset purchases brokered by a large firm have been puts or shorts.

Today, 631 trades were requested by clients, 435 of which were puts or shorts. Assume that today's

trades constitute a random sample of client requests to be made in the present and near future. At

5% significance, can we say that the proportion of puts and shorts among all trades has increased?

State the hypotheses, the p-value, and the conclusion.

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.