Sample size is 100 for each representative. 1.Construct a 95% confidence interva

Question

Sample size is 100 for each representative.

1.Construct a 95% confidence interval estimate for the difference in mean profits between sales representatives 1 and 2. In your answer discuss the relevant CI estimator and its conditions and provide evidence to support any assumptions madewhen constructing the estimate.
Define profits of all policies of sales representative 1 as population 1 and profits of all policies of sales representative 2 as population 2.  

2. Suppose that the two sales representatives for who the manager has data (rep 1 and rep 2) are the most successful ones in the team. The manager would like to award a prize to Rep 1 as he is more senior one of the two, but has been told that the prize should be awarded to the Rep that has obtained higher mean profits on all of their policies. The manager only has access to data for the 100 most recently matured policies of each sales representative as provided in the excel file. Using these data, conduct a hypothesis test at the 5% level to provide the manager with statistical evidence to decide whether sales representative 1 should be awarded the price.

3. Compute the p-value for the test you conducted in Q2. How strong is the statistical evidence in support of the alternative hypothesis? Based on the p-value, would your conclusion in Q2 change if the test were conducted at the 1% or 10% significance levels? Now, suppose we want to conduct a two-sided test that mean profits of Rep 1 and Rep 2 are different. Compute the p-value for such a two-sided test and interpret the result.  

4. The company has a new CEO that promised the shareholders a sales strategy that focuses on delivering profitable policies. Combining the profits of both representatives, 175 policies out of 200 are profitable. Compute a 90% interval estimate for the proportion of profitable policies, showing all your workings.   

5. Suppose the new CEO has formed a focus group that has set a target that over 80% of matured policies should have a positive profit. Conduct a hypothesis test at the 1% significance level to test whether the company currently fulfils this target.

Representative 1 Representative 2 Population STDEV 1300 Population STDEV 1300 Mean 1476.711 Mean 1591.529 Median 1433.588 Median 1587.134 Sample STDEV 1363.291 Sample STDEV 1362.537 Sample Variance 1858561 Sample Variance 1856507 Range 6825.534 Range 7635.692 Minimum -2512.71 Minimum -1681.31 Maximum 4312.82 Maximum 5954.378 CV 92% CV 86%

Explanation / Answer

Q1.
TRADITIONAL METHOD
given that,
mean(x)=1476.711
standard deviation , 1 =1300
population size(n1)=100
y(mean)=1591.529
standard deviation, 2 =1300
population size(n2)=100
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2   = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((1690000/100)+(1690000/100))
= 183.8478
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 183.8478
= 360.3416
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (1476.711-1591.529) ± 360.3416 ]
= [-475.1596 , 245.5236]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=1476.711
standard deviation , 1 =1300
number(n1)=100
y(mean)=1591.529
standard deviation, 2 =1300
number(n2)=100
CI = x1 - x2 ± Z a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ ( 1476.711-1591.529) ±Z a/2 * Sqrt( 1690000/100+1690000/100)]
= [ (-114.818) ± Z a/2 * Sqrt( 33800) ]
= [ (-114.818) ± 1.96 * Sqrt( 33800) ]
= [-475.1596 , 245.5236]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-475.1596 , 245.5236] contains the difference between
true population mean U1 - U2
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the difference between
true population mean U1 - U2
3. Since this Cl does n't contain a zero we can conclude at 0.05 true mean
difference is not zero

Q2.
Given that,
mean(x)=1476.711
standard deviation , 1 =1300
number(n1)=100
y(mean)=1591.529
standard deviation, 2 =1300
number(n2)=100
null, Ho: u1 > u2
alternate, H1: 1 < u2
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=1476.711-1591.529/sqrt((1690000/100)+(1690000/100))
zo =-0.62
| zo | =0.62
critical value
the value of |z | at los 0.05% is 1.645
we got |zo | =0.625 & | z | =1.645
make decision
hence value of | zo | < | z | and here we do not reject Ho
p-value: left tail - Ha : ( p < -0.62 ) = 0.26614
hence value of p0.05 < 0.26614,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 > u2
alternate, H1: 1 < u2
test statistic: -0.62
critical value: -1.645
decision: do not reject Ho
p-value: 0.26614
we have statistical evidence to decide sales representative 1 should be awarded the price.

Q3.
For 1% significance level,,p-value: left tail - Ha : ( p < -0.62 ) = 0.26614
hence value of p0.01 < 0.26614,here we do not reject Ho

For 10% significance level, p-value: left tail - Ha : ( p < -0.62 ) = 0.26614
hence value of p0.1 < 0.26614,here we do not reject Ho

Example 4.
given that,
possibile chances (x)=175
sample size(n)=200
success rate ( p )= x/n = 0.875
CI   = confidence interval
confidence interval = [ 0.875 ± 1.645 * Sqrt ( (0.875*0.125) /200) ) ]
= [0.875 - 1.645 * Sqrt ( (0.875*0.125) /200) , 0.875 + 1.645 * Sqrt ( (0.875*0.125) /200) ]
= [0.837 , 0.913]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 90% sure that the interval [ 0.837 , 0.913] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion

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