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Statistic Essay Example
 Category:Mathematics
 Document type:Math Problem
 Level:High School
 Page:1
 Words:638
Statistics for Business Assignment
Unit Name: Statistics for Business
Unit Number: 200032
2.1 C since no mention is made of gender, this is a marginal probability, the total who would
Buy divided by the total = 64/100 = 0.64
2.2 C Used the formula P (R∩T) =P(R ) * P (T/R)
2.3 D what is the probability of a randomly selected individual being a male who would buy?
This is just a joint probability. The number of «Male and would buy» divided by the total
= 28/100 = 0.28
2.4 D Used the formula P(R/T) =P(R and T ) / P (T)
2.5 D Used the formula P (Rc ∩T) =P(R ) * P (Rc)
• 50% of those surveyed are male (R), = 20
• 40% of the males surveyed would buy the product (T / R), =11
• 50% of the females surveyed and would buy the product (T /Rc ). =18
2.6 A since no mention is made of gender, this is a marginal probability
2.7 C Used the formula P(R/T) =P(R and T) / P (T)
SECTION B


p(x)=

5 minutes=1.5
15 minutes=4.5
So µ=4.5
P (x≤5) =
=0.17083
P (x≥3) =
mean=$ 40
P(X˃45)
Z=
Z=
Z=0.5 i.e. P(X˃45) is 0.5 standard deviation above the mean.
NB: if X is greater than mean we use the formula
X˃µ= 0.5+z
=0.5+0.1915(value from the Z table)
Z=
Z=
Z=0.5 i.e. P(X≥45) is 0.5 standard deviation above the mean.
X˃µ= 0.5+z
=0.5+0.1915(value from the Z table)
To construct a 90% confidence interval to estimate the proportion
of spam emails that get through.
Taking our α = 0.1, the area under the normal curve that is equal to (1 — α) or 90% is 0.697852
There for 90% C.I is
µ ± 0.697852
3±0.697852
Z ^{2 }* (p) * (1p) 

(e.g., .04 = ±4)
c = confidence interval, expressed as decimal
(.5 used for sample size needed)
p = percentage picking a choice, expressed as decimal
Z = Z value (e.g. 1.96 for 95% confidence level)
1.96 ^{2 }* 0.5 * (10.5) 


Year 1999
Sample size=17
Sample average= 170
Standard deviation= 19
Year 2006
Sample size=15
Sample average= 195
Standard deviation= 16
Assume that per diem cost are normally distributed and the population variance for 1999 and 2006 are equal
We use T test
The Difference between the mean of year 1999 and year 2006
= 170195
95% confidence interval of this difference is from 37.78 to 12.22
In ttest the two tailed p value equals 0.0004.
By conventional criteria , this difference is considered to be extremely statistically significant.[ CITATION SHI04 l 2057 ]
[ CITATION SHI04 l 2057 ]
T=
ɗɟ= 17+152
t= 3.9955
standard error difference 6.257

Non Defective
Defective
NB: No relationship between row and column frequencies.
Data Contingency table 

Non Defective Defective 

Expected Contingency table 

Non Defective Defective 

Observed(O) 
Expected(E) 
( OE)SQRUARED 

Day Defective 

Evening Defective 

Night Defective 
=6.05
Degrees of freedom =2
The table of= 0.9753
Since the computed
at 95% C.I is 6.05 and greater than 0.9753 tabulated
,there is no significant difference in defectives.


Regression equation SR= 184.5287AE — 593.955

The value of the slope is 184.5287 which is the change in the weekly sales revenue.

The experimental p value = 0.000 and the true value 0.001 thus there is a positive slope.

The coefficient of correlation is 0.9577121 and this value estimates the goodness of fit. It gives the percentage variation of the data explained by fitted line.

Predicted weekly sales
SR= 184.5287AE — 593.955
= 184.5287*10000593.955
=1844693.05(in $ 000s)
The excel program tested it at 95 % C.I
i) Calculate the un weighted aggregate price index for the year 2008
Solution
I=x100
I=x100

quantity
quantity
Batteries
Solution

batteries
Ip =x100
Ip1988,2008=x100
Ip1988,2008=194.18
REFERENCES:
SHIRLEY DOWDY, S. W. (2004). STATISTICS FOR RESEARCH. NEW JERSEY: JOHN WILEY AND SONS.