Management Accounting Essay Example

Running Header: Business Statistics

Management Accounting

City and State Where Institution is Located

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

	Frequency
6.50- 6.69
6.60-6.69
6.70-6.79
6.80-6.89
6.90-6.99
7.00-7.09
7.10-7.19
7.20-7.29
7.30-7.39
7.40-7.49
7.50-7.59
7.60-7.69
7.70-7.79	0
7.80- 7.89
7.90-7.99

The histogram

no. of customers
			Spending limit

1. 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.

1. The probability that the viewer is aged 18-34 is 0.69

b) The probability that a viewer is watching TV videos is 0.48

c) The percentage of viewers who are 18-34 and prefer watching user created videos is 39%

d) The percentage of viewers aged 18-34 who prefer watching user –created videos is 39%

e) The percentage of viewers aged 35-54 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^thMedian = 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^thMedian = 5

Mode = bi-modal = 65 and 79

Strength

Most frequent thus represents the general performance of the class

Weakness

Does not represent the whole population since bi-modal

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 = Bi-modal = 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^stIn most cases, the mean and the median agree as seen in the 1

t= 203-200/ (8/4) = 3/2 = 1.5 p= 0.3743

H0: μ = 200With a test statistic of 1.5 and p-value of 0.3743, we accept the null hypothesis

t= 198-200/ (5/5) = -2/1= -2 p =0.2952

H0: μ = 200With a test statistic of -2 and p-value of 0.2952, we accept the null hypothesis

t = 205-200/8/6 = 5/ 1.333 = 3.75 p =0.1659

H0: μ = 200With a test statistic of 3.75 and p-value 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= x-µ/s

=232,000-232,000/7,000 = 0

The probability is 0.5

1. Between 232,000 and 239,000 barrels?

232,000-239,000/7000 = 1

The probability is 0.8413

2. Less than 239,000 barrels?

The probability = 1-0.8413 = 0.1587

3. Less than 245,000 barrels?

Z= 232,000-239,000/7,000 = 1.86

Probability = 0.9686

4. More than 225,000 barrels?

Z = 232,000-225,000/7,000 =-1

Probability = 0.16

References:

, New York, John Willey & Sons.Business statistics: For contemporary decision makingBlack, K2017,