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Management Accounting Essay Example
 Category:Finance & Accounting
 Document type:Essay
 Level:High School
 Page:2
 Words:1122
1Statistics
Running Header: Business Statistics
Management Accounting
City and State Where Institution is Located

Frequency Distribution Table for average spending per customer (in dollars) at 74 Noodles & Company Restaurant

Frequency
6.50 6.69
6.606.69
6.706.79
6.806.89
6.906.99
7.007.09
7.107.19
7.207.29
7.307.39
7.407.49
7.507.59
7.607.69
7.707.79
0
7.80 7.89
7.907.99
The histogram


no. of customers 

Spending limit 


The Histogram
Above is the histogram representing the customers spending? The general appearance of the histogram is like a mountain with the majority of spenders being in the middle and hence the histogram is highest in the middle and the lowest at both sides. This is because few spenders spend too little while few spenders also spend much but a majority of spenders are average spenders as the shape and behaviour of the histogram indicates. However, the general skew is to the right meaning that majority of customers do not spend much.
How I choose the number of bins and the bin limit
I looked at the data provided and discovered that the lowest spender spend $6.54 with the highest spending $7.97. Thus, I decided that the lowest limit would have to start from 6.50 with each interval being $0.10 based on the level of spending per customer. With lowest interval starting at $6.50, the next interval had to be $6.59 and so on until $7.99 in order to cater for the highest spender. This resulted in a total of fifteen bins.
Histogram using more bins

Lower Limit
Upper Limit
Frequency (f)
Cumulative Frequency
Percentage
Cumulative percentage
0
0
0
0
0
0
0
0
0
0
0
0
Histogram from the frequency distribution table


No. of Customers 

Level of spending in $ 

On using more bins and differing bin limits, my visual impression of the data does not change since the general shape of the histogram remains the same as it is an indication of the customers’ spending behaviour described above.

The probability that the viewer is aged 1834 is 0.69
b) The probability that a viewer is watching TV videos is 0.48
c) The percentage of viewers who are 1834 and prefer watching user created videos is 39%
d) The percentage of viewers aged 1834 who prefer watching user –created videos is 39%
e) The percentage of viewers aged 3554 or prefer user created videos = 10+52 = 62%
3. Quiz 1 mean = 60+60+60+60+71+73+74+75+88+99 = 720/10 = 72 marks
Strength
The mean is representative of the whole population
Weakness
No body scores the mean mark
= 73, median = (71+73)/2 = 72 marks^{th} = 71, 6^{th}Median = 5
Strength
Same as mean and hence representative of the whole population
Weakness
Nobody has scored the median mark
Mode = 60 (Most frequent)
Strength
It is the most frequent meaning majority of class are in that level of knowledge
Weakness
Too low and does not compare with other measures
Quiz 2 mean = 65+65+65+65+70+74+79+79+79+79 = 72 marks
Strength
The mean is representative of the whole population
Weakness
No body scores the mean mark
=70+74= 144/2 = 72 marks^{th}Median = 5
Mode = bimodal = 65 and 79
Strength
Most frequent thus represents the general performance of the class
Weakness
Does not represent the whole population since bimodal
Quiz 3 mean = 66+67+70+71+72+72+74+74+95+99 = 76
Strength
Representative of the whole population
Weakness
Nobody scores the mean mark
Median = 72+72 =144/2 = 72 marks
Strength
Close to the mean hence representative of the whole population
Weakness
Lower than the mean
Mode = Bimodal = 72 and 74
Quiz 4 mean = 10+49+70+80+85+88+90+93+97+98 = 76
Strength
A bit high
Weakness
Not representative of the whole population since some students scored very low marks while others scored very high marks
Median = 85+88 = 86.5 marks
A bit high and represents the performance of majority of the class
Weakness
Too far from the lowest performance
Mode = No mode since no repeating number
Whether the measures of center agree
quiz. However, the mode in most cases does not agree with the median and mean since in some cases, there are two modes when two numbers repeat themselves and in some cases there is no mode. The reason is because the mode depends on the arrangement of the numbers and is coincidental since it depends on whether two or more people got the same marks. However, the median and the mean will agree in most cases, the mean will be found in the middle where the median also is. ^{nd} and 2^{st}In most cases, the mean and the median agree as seen in the 1
t= 203200/ (8/4) = 3/2 = 1.5 p= 0.3743
H0: μ = 200With a test statistic of 1.5 and pvalue of 0.3743, we accept the null hypothesis
t= 198200/ (5/5) = 2/1= 2 p =0.2952
H0: μ = 200With a test statistic of 2 and pvalue of 0.2952, we accept the null hypothesis
t = 205200/8/6 = 5/ 1.333 = 3.75 p =0.1659
H0: μ = 200With a test statistic of 3.75 and pvalue of 0.1659, we accept the null hypothesis
5. Mean score
Exam 1 = 81+79+88+90+82+86+80+92+86+86 = 850/10 =85
Exam 2 = 87+76+81+83+100+95+93+82+99+90 = 886/10 = 88.6
Exam 3 = 77+79+74+75+82+69+74+80+74+76 = 760/10 = 76
Standard error
Exam 1 = SE = s / sqrt( n ) = 4.15/sqrt 10 = 4.15/3.16 = 1.31
Critical value = 1.96
Margin of error = Critical value * Standard error = 1.96 * 1.31 = 2.57
95 percent confidence interval
2.57+ Exam 1 = 85
At 95% confidence level, the mean for exam 3 lies between 87.57 and 82.43 marks
Standard error = 7.71/sqrt 10 = 7.71/3.16 = 2.44
Critical value = 1.96
Margin of error = Critical value * Standard error = 2.44*1.96 = 4.78
95 percent confidence interval
4.78+ Exam 2 = 88.6
At 95% confidence level, the mean for exam 3 lies between 83.82 and 93.38 marks
Standard error = 3.52/3.16 = 1.11
Critical value = 1.96
Margin of error = 1.11* 1.96 = 2.18
95 percent confidence interval
2.18+Exam 3 = 76
At 95% confidence level, the mean for exam 3 lies between 78.18 and 73.82 marks
b) The three confidence intervals do not overlap
c) By comparing the three confidence intervals, I infer that the three means are not significantly different.
6. What is the probability of producing at least 232,000 barrels?
Z= xµ/s
=232,000232,000/7,000 = 0
The probability is 0.5

Between 232,000 and 239,000 barrels?
232,000239,000/7000 = 1
The probability is 0.8413

Less than 239,000 barrels?
The probability = 10.8413 = 0.1587

Less than 245,000 barrels?
Z= 232,000239,000/7,000 = 1.86
Probability = 0.9686

More than 225,000 barrels?
Z = 232,000225,000/7,000 =1
Probability = 0.16
References:
, New York, John Willey & Sons.Business statistics: For contemporary decision makingBlack, K2017,