Executive Summary:
Par, Inc has developed a new coating designed to resist cuts and provide a more durable ball. One concern for Par, Inc was the effect of the new coating on driving distances. Par would like the new cut-resistant ball to offer driving distances comparable to those of the current-model golf ball. To compare the driving distances for the two balls, 40 balls of both new and current models were subjected to distance test. The testing was performed with a mechanical hitting machine so that any difference between the mean distances for the two models could be attributed to a difference in the two models.

Based on the sample studied of 80 golf balls both current and new models, I have done the following analysis: 1. Formulated a hypothesis test for Par’s comparison of the driving distances of the current and new golf balls. 2. Analyzed data to provide a hypothesis testing conclusion and recommendation. 3. Provided descriptive statistical summaries of the data for each model 4. Calculated 95% confidence interval for the population mean of each model, and the difference between the means of the two populations. 5. Discuss if a larger sample size was needed with the golf balls

First off, I’ve conducted an hypothesis test for Par, Inc comparison and the null/alternative hypothesis are as follow:

H0: µ1 - µ2 = 0 (they are the same)
Ha: µ1 - µ2 ≠ 0 (they are not the same)

The test I will be conducting is a two-tail test.

The P-Value for this test is .01836 and because it is greater than alpha of .05, thus we do not reject the null hypothesis. My recommendation to Par, Inc is to put the new coating for the golf balls to production.

Descriptive Statistic for both models are as follows:

...This report is case study of Par, Inc.
Par, Inc., is a manufacturer of golf equipment. The company wants to check whether this new type of golf ball is comparable to the old ones. Therefore they did a test with 40 samples of both new balls and old ones. The testing was performed with a mechanical hitting machine so that any difference between the mean distances for the two models could be measure to evaluate the difference in the two models, presented with the sample size of 40 and confidence level of 95%.
The table below depicts a descriptive statistical summary for the case study.
| Current Model | New Model |
Sample Mean | 268.89 | 272.98 |
Standard Error | 1.30 | 1.55 |
Standard Deviation | 8.20 | 9.81 |
Standard variances | 67.28 | 96.28 |
Sample Size | 40 | 40 |
Confidence Level (95%) | 2.62 | 3.14 |
The hypothesis for Par, Inc. to compare the driving distance of the current and new golf balls is
H0 : µ1 - µ2 = 0 (they are the same)
Ha : µ1 - µ2 ≠ 0 (the are not the same).
P-Value for this two tailed test is 0.06, which is greater than level of significance α (0.05).
Hence, H0 will not be rejected which shows that Par, Inc. should take a new ball in production as the P value indicates that there is no significant difference between estimated population mean of current as well new sample model....

...Case #1
Par, Inc.
Date: October 8th 2011
Question 1
Formulate and present the rationale for a hypothesis test that Par could use to compare the driving distances of the current and new golf balls
Ans:
The rational is that Par, Inc. is concerned with making sure that the market share or the new ball is comparable to the previous with the new cut-resistant, longer lasting ball. We need to compare the mean driving distance in order to be certain that the new ball efforts are worth it. Therefore we need to formulate a hypothesis test.
The text tells us that Par, Inc. would like to sell the new ball once it’s comparable to the previous. Therefore:
Ho: μ₁ -μ₂ ≥ 0 the mean driving distance of the new balls is greater than or equal to that of the old balls
Ha: μ₁ - μ₂ ‹ 0 the mean driving distance of the new balls is less than that of the old balls,
α = 0.05
CURRENT | NEW |
270.28 | 267.50 |
8.75 | 9.90 |
40 | 40 |
Mean
Sample Standard Deviation
N
t= (x̄₁ - x̄₂) – (μ₁ - μ₂)0
√Sp²[(1/n₁) + (1/n₂)]
Sp² = (n₁-1) s₁² + (n₂-1) s² = (40-1)(8.75) +(40-1)(9.90) ² = 6808.23 = 87.285
n₁ + n₂ - 2 40+40-2 78
Sp² = 87.285
Df= 78
t=(270.28 – 267.5) – 0 = 1.33...

...CaseProblem: Par, Inc.
Section I: Summary
Par, Inc., a major manufacturer of golf equipment believes that a cut-resistant, longer lasting golf ball could increase their market share. In addition to the requirement that the ball be longer lasting, they wanted to ensure that the new coating would not reduce driving distances, and would be comparable to the current product.
Section II: Relevant Statistical Results
Statistic
Current Model
New Model
Sample Mean
270.275
267.500
Standard Error
1.3840
1.5648
Standard Deviation
8.7530
9.8969
Sample Size
40
40
Confidence Level (95%)
2.62
3.14
Degrees of freedom
39
39
t = 1.33
Comparisons
1. A two tail hypothesis test was conducted based on the sample studies of 40 current and 40 new golf balls.
The testing was performed with a machine designed to hit the ball the same every time. Both samples were tested with the same machine to ensure consistent strikes.
2. The hypothesis for Par, Inc. to compare the driving distances of the current and new golf balls is:
H0: µ1 - µ2 ≥ 0 (mean distance of new greater than or equal to the old)
Ha: µ1 -µ2 < 0 (mean distance of new less than old)
3. Statistical basis
n1 = 40 n2 = 40
x1 = 270.28 x2 = 267.5
s1 = 8.75 s2 = 9.90
α = .05
df = 78
t = (270.28 – 267.5) – 0 = 1.33
P-value = .10
P-value > .05 (or α); do not reject H0
4. The 95% confidence interval for the...

...The management at ParInc. believes that with the introduction of a cut-resistant, longer-lasting golf ball could increase their market share. A new golf ball coating designed to resist cuts and provide a more durable ball have been developed and tested. A sample of 40 balls of both the new and current models were tested with a mechanical hitting machine so that any difference between the mean distances for the two models could be attributed to a difference in the two models. Therefore, the hypothesis test that Par could use to compare the driving distances of the current and new golf balls can be formulated as follows:
It is set to be a two-tailed test which refers to the difference between the mean distances. Here, refers to the mean driving distance of the new golf ball whilst refers to the mean driving distance of the current golf ball. When the difference between the mean distances is equal to 0, therefore, the null hypothesis rejected. This shows that the new golf ball is better.
With the formula, the value of the test statistics was computed. This is a necessary step to find the p-value. Next step would be computing the degree of freedom using the formula as such: . The findings presented a test statistics of and the degree of freedom of 76. Therefore, the p-value is greater than the level of significance which was chosen to be 0.05 and the null hypothesis is not rejected. It is recommended for...

...Here is the report about Par, Inc., is a major manufacturer of golf equipment test whether the new ball drive longer distance than the current model. To compare the driving distances for the two balls,
40 simple tests both of new and current models were subjected to distance tests. According to the data, we got the information we need for a hypothesis test as follow:
| Current | New |
Means | 269.42 | 266.67 |
Count | 40 | 40 |
Standard Deviation | 8.09 | 9.79 |
Confidence Level(95.0%) | 2.59 | 3.13 |
| | |
The 40 simple of both current and new model golf balls show that the average distance of the new ball drive is less than the current model, but the standard deviation of new ball is 9.79 which is larger than the current one and It imply the new ball is not stable as the current one, it has more chance drive longer or shorter than the current model.
But at present, we can’t get the conclusion which ball drive longer distance, So we need a hypothesis test the difference of this two ball models. The hypothesis test suggested follows:
H0:µ 1-µ 2≤0
Ha:µ 1-µ 2>0
We use this formula for a hypothesis test to compare the driving distance of the current and new golf balls. After analyses, we get the conclusion that we can’t reject the null hypothesis. Because this is a hypothesis test about two different populations, standard deviation of population is unknown and we use t-test and the p-value we...

...BLADE INC. CASE
1. What are the advantages Blades could gain from importing from and/or exporting to a foreign country such as Thailand? Ans: The advantages Blades could gain from importing from and/or exporting to Thailand could be Decrease their cost of goods sold, and increase Blades’ net income since rubber and plastic are cheaper when imported from a foreign country such as Thailand. Due to its superior production process Thai firms could not duplicate the high-quality production process , so establishing a subsidiary in Thailand would preserve blade sales before Thai competitors. Allow Blades to explore the option of exporting to Thailand by building relationships with some local suppliers. As far as exporting is concerned, Blades could become the first firm to seller roller Blades in Thailand. Diversify their investment by opening option to export to other countries beyond Thailand to ensure company sustainability.
2. What are some of the disadvantages Blades could face as a result of foreign trade in the short run? In the long run?
Ans: The disadvantages Blades could face as a result of foreign trade in the short run are: Exchange rate risk. Blades would be exposed to currency fluctuation in the Thai baht if importation cost increase without Thai suppliers adjusting their price. International economic condition; if Thailand’s economy undergoes recession, Blades would suffer from sales decrease in Thailand. In the long run,...

...
Introduction
The purpose of this report is to show if the new golf ball coating design does induced resist cuts and provide a more durable ball.
Hypothesis
µ1 is the mean driving distance of currently produced golf balls
µ2 is the mean driving distance of newly designed golf balls
Null hypothesis (H0): µ1- µ2 ≤ 0
Alternate hypothesis (Ha): µ1- µ2 > 0
Analysis
The sample mean for current golf balls is 270.275, while the new golf balls have a sample mean of 267.50. So on average the current golf balls have an advantage of 2.775 yards. Using a 5% significance level, a p value of 0.094 is obtained. As this is larger than the significance level, the null hypothesis cannot be rejected. The conclusion is that this data does not provide statistical evidence that the new golf balls have a lower mean driving distance.
However, this does not mean that the new golf balls do not have a better driving distance only that this set of data does not support it. A type II error could have been made, where we have not rejected the null hypothesis, but it is in fact false. Only more data can help give a better idea on whether there is a difference in the driving distance of the current and new golf balls. So it is recommended that more data be acquired and a new test be performed with a bigger sampling size.
Descriptive Summaries
Confidence Intervals (95%)
Confidence interval at 95% for the population mean driving distance of current golf balls.
The 95%...

...CaseProblem: Blades, Inc.
1. One point of concern for you is that there is a tradeoff between the higher interest rates in Thailand and the delayed conversion of baht into dollars. Explain what this means.
ANSWER: If the net baht-denominated cash flows are converted into dollars today, Blades is not subject to any future depreciation of the baht that would result in less dollar cash flows.
2. If the net baht received from the Thailand operation are invested in Thailand, how will U.S. operations be affected? (Assume that Blades is currently paying 10 percent on dollars borrowed, and needs more financing for its firm.)
ANSWER: If the cash flows generated in Thailand are all used to support U.S. operations, then Blades will have to borrow additional funds in the U.S. (or the international money market) at an interest rate of 10 percent. For example, if the baht will depreciate by 10 percent over the next year, the Thai investment will render a yield of roughly 5 percent, while the company pays 10 percent interest on funds borrowed in the U.S. Since the funds could have been converted into dollars immediately and used in the U.S., the baht should probably be converted into dollars today to forgo the additional (expected) interest expenses that would be incurred from this action.
3. Construct a spreadsheet that compares the cash flows resulting from two plans. Under the first plan, net baht-denominated cash...

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