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# Statistics for Business Assignment Example

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## Table of Content

### Solution to Question 1

#### Part a)

Mean for Quiz 1 = 60 + 60 + 60 + 60 + 71 + 73 + 74 + 75 + 88 + 9910

=72

Median for Quiz 1 = 71+732

=  72

Mode for Quiz 1 = most repeated value = 60

Mean for Quiz 2 = 65+65+65+ 65+ 70+ 74+79+ 79+ 79+ 7910

=72

Median for Quiz 2 = 70+742

=  72

Mode for Quiz 2 = most repeated value = 65, 79

Mean for Quiz 3 = 66+ 67+ 70+ 71+ 72+ 72+ 74+74+ 95+ 9910

=76

Median for Quiz 3 = 72+722

=  72

Mode for Quiz 3 = most repeated value = 72, 74

Mean for Quiz 4 = 10+ 49+ 70+ 80+ 85+ 88+ 90+ 93+ 97+ 9810

=76

Median for Quiz 4 = 85+882

=  86.5

Mode for Quiz 4 = most repeated value = No mode

#### Part b)

No, these measures do not agree because they did not give the same value for each quiz.

#### Part c)

• When the mean and median do not agree with each other, then the shape of the distribution should be looked at to determine the appropriate measure.

• For quantitative data, the reliable measure of central tendency is the mode.

• When the mean and median agree with each other, the shape of the distribution can be looked at to determine the more appropriate measure.

• For quantitative data, the unreliable measure of central tendency is the mode.

#### Part d)

• Data is symmetric for Quiz 1 as mean = median

• Data is symmetric for Quiz 2 as mean = median

• Data is skewed in the right direction for Quiz 3 as mean > median

• Data is skewed in the left direction for Quiz 4 as mean < median

#### Part e)

The student's performances are betters as an average in quiz 3 and quiz 4 compared to quiz 1 and quiz 2. ​​

### Solutions to Question 2

#### Part a)

Sample = N = 100

almonds= x= 19

Sample proportion = P =xN

=19100

= 0.19

sample proportion is distributed normally, onfidence interval at 90%:

P ± Std Error

Std Error = 1.645 0.19 0.81100

Std Error = 0.0645

90% confidence interval = 0.19 ± 0.0645

= 0.2545 & 0.1255

#### Part b)

np=(100)( 0.19)

= 19>10

n (1-p)= 100 (1  0.19)

= 100 (0.81)

= 81>10

Hence,normality can be assumed.

#### Part c)

e=0.03

z=1.645

n = ze2pq

= 1.6450.0320.19)(0.81

= 462.65  4.63

#### Part d)

Testing will reveal it to the manager that either the product is according to its advertisement or not; he or she will measure the quality or limit through the sampling technique.

### Solution to Question 3

#### Part a)

p value=0.6019

at α = 0.05, t Stat table value = - 1.706

calculated t Stat value = - 0.528

table value &lt; t stat and p value &gt; α

p-value> α

0.6019>0.05

and table value<t Stat

- 1.706< -0.528

yes, the coefficient of the price is significantly different from 0 at α=0.05

#### Part b)

The value of R2 tells that the independent variable, i.e., defines the dependent variable, i.e., perceived sound quality by 0.01104. It is the coefficient of determination.

#### Part​​ c)

The higher price does not imply higher sound quality, and the negative sign with price shows the negative relationship between variables.

### The​​ Solution to Question 4

#### Part a)

Null Hypothesis:

H 0 : μ = 2 days

Alternative Hypothesis:

H 1 : μ > 2 days

#### Part b)

Error type = Type II Error

Impact: the package will not be delivered as scheduled or promised​​

#### Part c)

Error type = Type I Error

Impact: the company’s credibility will be needlessly lost, and the responsible person would have to face serious consequences.

#### Part d)

Worse error from the company's standpoint = Type I error

Reason: the company cannot stand with bad credibility or reputation in the market.

#### Part 2)

Worse error from the consumer’s standpoint = Type II error

Reason: They would never want their package to be delivered late.

#### References

• Doane, 2006.​​ Applied Statistics In Business And Economics.​​ S .l.: Tata McGraw-Hill Education.

## Assignment 2

### Question 1:​​

Prof. Hardtack gave four Friday quizzes last semester in his 10-student senior tax accounting class.

Quiz 1:​​ 60, 60, 60, 60, 71, 73, 74, 75, 88, 99

Quiz 2:​​ 65, 65, 65, 65, 70, 74, 79, 79, 79, 79

Quiz 3:​​ 66, 67, 70, 71, 72, 72, 74, 74, 95, 99

Quiz 4:​​ 10, 49, 70, 80, 85, 88, 90, 93, 97, 98

• ​​ Find the​​ Mean,​​ Median, and​​ Mode for​​ Each​​ Quiz.​​

• Quiz 1

Mean = 72

Median = 72

Mode = 60

• Quiz 2

Mean = 72

Median = 72

Mode = 65

• Quiz 3

Mean = 76

Median = 72

Mode = 72

• Quiz 4

Mean = 76

Median = 86.5

Mode = no mode

• Do these​​ Measures of​​ Center​​ Agree? Explain.

The mean shows each student's average score, i.e., if each student had scored the same in the quiz, the mean would have represented their score. Moreover, the mean represents the average value showing that 5 (half) students have scored higher in the quiz than the mean value, and the other 5 students have scored lower than the mean value, and the same goes for median and mode.

In quiz 1 and quiz 2, the mean and median agree as the value is the same for both center measures while mode does not agree to mean and median. In quiz 3, median and mode agree as the value is the same for both measures while mean does not agree to median and mode. In quiz 4, none out of mean, median, and mode agree to each other.

• ​​ For Each Data Set, Note​​ the Strengths​​ or Weaknesses​​ of Each Statistic Center.

For each data set, the model has been the weakest as the central tendency can be achieved through the small test set’s randomness. The students' grades vary in the date sets aside from some values, then the similar values can be put at the top or bottom of the data set. Each data set's risk can be high if there are so many good grades or so many bad grades. The strength of these measures is that the grades of students are in between—the strength of the median in that the grades are spread feasibly. Mean seemed to be a more strengthened measure and best choice to some extent as well as it has provided the best average values that can be easily compared.​​

• Are​​ the Data Symmetric​​ or Skewed? If Skewed, Which Direction?​​

For quiz 1 and quiz 2, the data is symmetric because the mean and median are equal. The data for quiz 3 is skewed in the right direction because the mean is more significant than the median, while data for quiz 4 is skewed in the left direction because the mean is less than the median.

• Describe​​ and Equate​​ the Results​​ of Students​​ on Each Questionnaire Momentarily.

If we consider the mean value, students have scored higher and equal in quiz 3 and 4 while they have scored lower and equal in quiz 1 and 2. If we consider the median value, students have scored higher on quiz 4 and equal in quiz 1, 2, and 3. If we consider the value of mode, then students' performance regarding quiz 4 remains unknown, and students have scored higher on quiz 3 and lower in quiz 1.

### Question 2:​​

In a sample of 100 Planter's Mixed Nuts, 19 were found to be almonds.​​

• Describe and briefly compare the outcomes of students on each questionnaire.

Let’s assume:

Sample = N = 100 and y = 19

Sample proportion = P = y / N

= 19 / 100

= 0.19​​

Now we can assume that sample proportion is distributed normally, then the confidence interval at 90% is:

P + Standard Error , P – Standard Error

Standard Error =​​ 1.645 0.19 0.81100

= 0.0645

90% confidence interval = 0.19 + 0.0645 , 0.19 – 0.0645

= 0.2545 , 0.1255

• May normality be​​ Assumed? Explain.​​

Let’s check the normality:

y * P = 100 * 0.19

= 19 (that is greater than 10)

Similarly,

= 100 (1 – 0.19)

= 100 (0.81)

= 81 (that is greater than 10)

Hence, the normality has worked, so yes, it is possible that normality can be assumed.

• For 90 Percent Conviction​​ and An Error Of ±0.03, What Sample Size Will Be Required?

Error = e = 0.0.3 while z = 1645

Sample size = n =​​ ze2pq

=​​ 1.6450.0320.19*0.81

= 462.65​​ ​​ 4.63

• Why Would A Quality Control Manager​​ at Planter's Need To Understand Sampling?

The quality manager at the Planters needs to understand the sampling because they can check it through the testing that either the product is according to the advertisement. The quality manager will be able to calculate the limits of quality control through the sampling techniques.

### Question 3:​​

Consider the following Excel regression of perceived sound quality for 27 stereo speakers as a feature of the price.

 Regression​​ Statistics R Square 0.01104 Standard Error 4.02545 Observations 27

 Statistic Coefficients Std Error t Stat P-value Lower 95% Upper 95% Intercept 88.4902 1.67814 52.731 0.0000 85.0340 91.9464 Price −0.00239 0.00453 −0.528 0.6019 −0.01172 0.00693

• Is​​ the Coefficient Of Price Significantly Different From Zero At Α = .05?

The p value is 0.6019

t Stat table value = - 1.706 (at α = 0.05)

calculated t Stat value = - 0.528

now:

table value < t stat and p value > α

-1.706 < - 0.528 and 0.6019 > 0.05

It can be observed that at α = 0.05, the coefficient of price is different significantly.​​

• What does the R2 tell you?​​

The value of R2 tells the variability of the data that response is 1.01104 percent around its mean. Usually, the value of R2 lies between 0 and 100 percent.

• Given These Results, Would You Conclude That A Higher Price Implies Higher Sound Quality?

The higher price does not imply the higher, the higher the sound quality because the price's value is negative, showing the negative relationship between variables. ​​

### Question 4：

Suppose a pickup and delivery company states that their packages arrive within two days or less on average. You want to find out whether the actual average delivery time is longer than this. You conduct a hypothesis test.

• Set​​ Up the​​ Null and​​ Alternative​​ Hypotheses.

Null Hypothesis: H0: μ = 2 days

Alternative Hypothesis: H1: μ > 2 days

• Suppose You Conclude Wrongly That​​ the Company’s Statement About​​ the Average Delivery Time Is Within Two Days. What Type​​ of Error Is Being Committed,​​ and What Is​​ the Impact Of That Error?

The claim of the company is not rejected, although it would have rejected. The error is of Type II error, and the opportunity has missed saying that the company's claim was wrong. This error will be that the package will not be delivered to the people within 2 average days and could become dissatisfied eventually.​​

• Suppose You Conclude Wrongly That​​ the Delivery Company’s Average Time​​ to Delivery Is, In Fact, Longer Than Two Days.​​ What Type​​ of Error Did You Commit​​ and What Is The Impact Of This Error?

The claim of the company is rejected, although it should not have. Now, this error is of Type I error as the false alarm has caused. The impact of such a kind of error will be that the company's credibility will be lost needlessly, and some real problems could be caused to the individual responsible for such false alarm.

• Which Error Is Worse​​ from The Company’s Standpoint, A Type I Or A Type II Error? Why?

The company would eagerly want to minimize the Type I error chances as it would not want to lose the reputation unfairly. However, by minimizing this error, the chances of Type II error will increase. Still, Type I error is worse for the company.​​

• Which Error Is Worse​​ from A Consumer Standpoint, A Type I Or A Type II Error? Why?

From the consumer standpoint, the Type II error is worse because they want their package to be delivered at the time. However, a chance for the false alarm will increase if the consumer would minimize the Type I error.