# Statistic Essay Example

• Category:
Mathematics
• Document type:
Math Problem
• Level:
High School
• Page:
1
• Words:
638

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

1. 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)
1. 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.

1. Regression equation SR= 184.5287AE — 593.955

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

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

1. 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.

1. 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.