Couple of questions.. Need solving. (Attempts shown)

techguy

Hi,
Currently revising for a repeat exam and I am absolutely hopeless. Could somebody here give some help with these? Thanks.

Question 1:

Early network communication systems suffered from noise. In a particular system a 1 is
transmitted onethird
of the time and a 0 is transmitted twothirds
of the time. When a 0 is
sent the probability that it is received correctly is 0.85 and the probability that when a 1 is
sent it is received correctly is 0.9. Use Bayes' Theorem to find the probability that a 0 was
transmitted given that a 0 was received.

Attempt:

P(tran1)=1/3           P(receive1)=0.9
P(trab0)=2/3 <= [COLOR="Red"]A[/COLOR]   P(receive0)=0.85 <= [COLOR="Red"]B[/COLOR]

Find: P(A|B)     -> Use Bayes Theorem

P(A|B) = [U]            P(B|A) P(B)                  [/U]
             P(B|A)P(B) + P(A|&#172;B)P(&#172;B)

Sub in values but I can't seem to find P(B|A) at least when I do it equates to .85..

Question 2:

In October 2007, a sample of 2000 U.S. voters were asked if George W. Bush was doing a
good job as President of the U.S.A. 725 respondents thought the President was doing a good
job. From this information, infer the range for the population proportion that think he has
done a good job.

Attempt:

2000 Voters
725 = Good Job
1,272 = Bad Job

What is [I]Range for the population proportion[/I]?

Any help greatly appreciated.Thanks.

MathsManiac

First one:
Your problem is with your setting up: 0.85 is not P(received 0), it's P(received 0 | sent 0). Similarly for the 0.9.

Try that and see you get on.

Second one:

Your "point estimate" for the population proportion is the sample proportion: 725/2000, which is 36.25%.

The quick way to infer an estimated range for the proportion is to use the quick rule that the "margin of error" tion is 1/sqrt(n). In this case, 1/sqrt(2000) is 2.23%, so the estimated range is from 34% to 38.5%.

A more specific/accurate "inferred range" is to get the 95% confidence interval, which is: the estimate (+/-) 1.96*(standard error)

The standard error for a population proportion is given by: sqrt[p(1-p)/n].

If you work that out, you'll get a slightly narrower estimated range than the one given by the quick rule.

techguy

Ok,

I've attempted Bayes Theorem again:

P(s1) = 1/3 = 0.33
P(s0) = 2/3 = 0.66

P(r1 | s1) = 0.9                    p(¬r1 | s1) = 0.1   ==> p(r0 | s1)
P(r0 | s0) = 0.85                   p(¬r0 | s0) = 0.85  ==> p(r1 | s0)

P(s0 | r0) = [COLOR="Red"]??[/COLOR]


P(s0 | r0) = [U]           p(r0 | s0)p(s0)                   [/U]
                 p(r0 | s0)p(s0) + p(r0 | s1)p(s1)

===>   [U]        0.85(0.66)         [/U]
           0.85(0.66) + 0.1(0.33)

===> 0.944 ===> P(s0 | r0)

Here's my attempt at the second one..:

P = [U] 725 [/U]  ==> 0.3625
      2000

??? = 0.3625 +- 1.96 * SQRT(.3625*(1-.3625)/2000)

??? = 0.3625 +- 0.021068575

36.25 % +- 2.1%

Is this what I am supposed to do??  The values seem slightly off compared to yours?

techguy

Here's another Bayes Theorem question:

Question:
A company is about to launch a new drink on the market. In the past, 65% of its drinks have successfully launched.
Before any soft drink is launched, the company conducts research and receives a report predicting favourable or unfavourable sales.
In the past, 83% of successful drinks and 24% of the unsuccessful ones received favourable reports. What is the probability this new drink will be successful given that it receives a favourable report?


Answer:

P(success) = 0.65
P(fail) = 0.35

P(favourable | success ) = 0.83
P(favourable | fail ) = .24

P(success | favourable) = ??

P(success | favourable) = [U]            P(fav | sucess) P(success)         [/U]
                           P(fav | success)P(success)+P(fav | fail)P(fail)

==> [U]0.83(0.65)[/U]
0.83(0.65) + 0.24(0.35)

==> 86.52% chance of success given a favourable report.

Another One:

In a recent poll, 23% of a sample of 500 Irish drivers admit that they sometimes break the speed limit. From this, infer the percentage of the total population of drivers who break the speed limit.

Answer:

P = 23% N = 500

Pi (What am  I looking for?) = 0.23 +- 1.96 * SQRT(P*(1-P)/N)

==> 0.23 +- 1.96 * 0.018820201
==> 0.23 +- 0.0368876

23% +- 3.% of the total population break the speed limit.

I really appreciate you looking at these.. I have a repeat exam on tuesday, I have nobody else to look at these and I think i'm going to struggle to get 40% (Shame on me in the first place I know :o)

techguy

Here are some attempts on Hypothesis Testing.. I am fairly confident that my work is correct I just need some assurance..

Question 1:

A train timetable says that the time it takes to travel from Maynooth to Dublin Connolly is 25 minutes. A survey of the arrival times of six trains resulted in a mean of 20 minutes. Assuming that the time is normally distributed, and that the standard deviation is 7 minutes, perform a hypothesis test with a significance level of 0.05 to see if the claim is correct.

Attempt:

µ = 25                                     Stnd Dev = 7
Sample Mean ¬X = 20                 Alpha (Significance) = 0.05
Number of samples N = 6

Null Hypothesis H0 = µ => 25
Alt Hypothesis H1 = &#181; < 25

Test Stat Z:
=> [U]   &#172;x - &#181;0   [/U]
   StdDev/SQRT(N)
=> [U-5[/U]
     2.8577380
=> [COLOR="Red"]-1.7496[/COLOR]

Rejection Region : Z < - 1.96

Conclusion: Z is not less than -1.96 so the Null hypothesis is accepted. It takes on avg 25 minutes to get to Dublin from Maynooth.

Sorry if that one is a bit all over the place.

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Question 2:


A bank bases its sales strategy on the assumption that the mean size of a mortgage taken out by a customer is €300,000. A recent survey of five customers indicates that their mean mortage was €280,000 with a standard deviation of €25,000. Using α=0.05, test if the bank's assumption is correct.

Attempt:

µ = 300,000                                 Std Dev = 25,000
Sample Avg ¬X = 280,000              Significance (Alpha) = 0.05
Number of samples N = 5

H0 = µ => 300,000
H1 = &#181; < 300,000

Test Statistic Z:

=>[U]  &#172;X - &#181;0[/U]
StdDev/SQRT(N)

=> [U]-20,000[/U]
    11,180

=> [COLOR="Red"]-1.78891[/COLOR]

Rejection Region: Z < - 1.96

Conclusion: Test statistic Z is not less than -1.96 (Not in rejection region) so  accpet the null hypothesis. The bank is correct in their assumptions that the mean mortgage is 300,000