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Statistic Essay Example
- Category:Mathematics
- Document type:Math Problem
- Level:High School
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- 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 e-mails 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) * (1-p) |
|
(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 * (1-0.5) |
|
-
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
= 170-195
95% confidence interval of this difference is from -37.78 to -12.22
In t-test 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+15-2
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) |
( O-E)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*10000-593.955
=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.