A customer buys 1,000 shares of XYZZ stock in a margin account and pays the required 50% margin on settlement date. The customer
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Answer: Please refer to Explanation
Explanation:
When there are no or relatively low control procedures in a company, there is a threat of financial mismanagement and misdemeanors. This is why control procedures are needed, to address this and stop the leakage of company resources.
1. Failing to take available purchase discounts for prompt payment.
d. File invoices by due date.
e. Maintain a cash budget.
Here two things can be done to control the threat. Firstly, by paying invoices during the discount period, the company can be able to take discounts on goods and services provided to it. Also by maintaining a cash budget, a company can put when a payment is due to be able to claim a discount and act accordingly.
2. Recording and posting errors in accounts payable.
Conduct an automated comparison of total change in cash to total changes in accounts payable.
Using a program to check whether the amounts in the cash account corresponds to the payments on the Accounts payable account will tell you if the amounts tally and will therefore reduce errors.
3. Paying for items not received.
Issue checks only for complete voucher packages (receiving report, supplier invoice, and purchase order).
When issuing checks, make sure that all the above mentioned reports are in order. That way you can check if the goods were delivered as well as if they were even ordered properly in the first place.
4. Kickbacks.
Require purchasing agents to disclose financial or personal interests in suppliers.
Train employees in how to properly respond to gifts or incentives offered by suppliers.
By requiring that purchase agents disclose their relationships with suppliers, you can monitor to check and see if there is a possibility of kickbacks occuring.
Also, by training employees on acceptable methods of receiving gifts, they can know when it is no longer a gift but rather a kickback.
5. Theft of inventory.
b. Document all transfers of inventory. c. Restrict physical access to inventory.
By documenting all transfers going in and out of inventory, the true inventory figure can be known from the records and then used to match with the actual inventory to see if they truly tally.
Restricting the amount of people who have access to the inventory to a few trusted people also limits the amount of people who can steal the inventory as well as making it easier to find out who did when it is done because the focus can be on a few people.
a. The point estimate of the married people that sleep well is = 0.6299. The confidence interval for the mean hours of sleep for those that have never been married but report good sleep is 0.6186, 0.6412
b. The point estimate of the number of hours for people that have never been married = 7.2257
95% confidence interval for the mean number of hours of sleep for those who have never been married = 7.2065, 7.2447
c. the estimated number of sleep for those that have never been married = 7.729 hours
a. The point estimate
Those who have never been married n = 7044
Xp = those who sleep for 7 hours at least
To get Xp, the <u>excel function </u>
Countif(range, criteria)
Countif(A:A, >=7)
This gives Xp = 4437
Point estimate =
= 0.6299
The point estimate of the married people that sleep well is = 0.6299
At 95% interval of confidence,
α = 1-0.95 = 0.05
α/2 = 0.05/2 = 0.025
Z0.025 = 1.96 ( Use the standard probability normal table)
Margin of Error
=
= 0.0113
The 95% confidence interval
= 0.6299 - 0.0113 , 0.6299 + 0.0113
= 0.6186, 0.6412
The confidence interval for the mean hours of sleep for those that have never been married is 0.6186, 0.6412
b. Sample mean
x = ∑x/n
x = number of sleep of the never been married
Mean is calculated using excel function
Average(range)
Average(a:a)
This gives the sample mean = 7.2257
The point estimate of the number of hours for people that have never been married = 7.2257
<u>Standard deviation</u> using excel =
Stdev(A:A) = 0.8211
= 0.0192
The<u> confidence interval</u> =
= 7.2257-0.0192, 7.2257+0.0192
= 7.2065, 7.2447
95% confidence interval for the mean number of hours of sleep for those who have never been married = 7.2065, 7.2447
c. From the available data, the mean number of sleep per day that is gotten from the 4437 people that are unmarried and sleep healthily for at least 7 hours = 34295.1
Mean = 34295.1/4437
= 7.729
the estimated number of sleep for those that have never been married = 7.729 hours
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