Expert: Ahmed Salami - 11/4/2008

Question
Hello,


I pay my apologies for using short-cut language. I also thank you for replying so soon to my mail. I have 2 questions related to theory of probability. KIndly answer them.

 
1.    A central university has a student population of 60,000. The university is interested in determining what proportion of them is in favour of a new grading system. Determine a sample size with confidence level of 95% that will show the true proportion of population in favour of the new system within plus and minus 0.02.


2.     A telescope manufacturer wants its telescopes to have standard deviations in resolution to be significantly below 2 when focusing on objects 500 light-years away. When a telescope is used to focus on an object 500 light years away 30 times, the sample standard deviation turns out to be 1.46.

a. State explicit null and alternate hypotheses

b. Test your hypothesis at the α=0.01 level.  

Answer
Hi Nandita,
1)Applying the  general formula for a  confidence interval, the
confidence interval for a proportion is

p ± zσ

where p is the proportion in the sample, z depends on the level of
confidence desired, and σ is the standard error of a proportion.

σ = sqrt[p(1-p)/N]

Since we dont know how many are in favour of the new grading system,
we take p = 0.5 for our estimation of σ.

Also, we require that p ± zσ = p ± 0.02
zσ = 0.02

A z table can be used to determine that z for a 95% confidence
interval is 1.96
And so,
1.96σ = 0.02
σ = 0.01
sqrt[0.5(1 - 0.05)/N] = 0.01
sqrt[(0.5)(0.5)/N] = 0.01
0.5/sqrtN = 0.01
sqrtN = 0.5/0.01
sqrtN = 50
N = 2500

2)
a) The Null Hypothesis
 H0: S = 2   (S represents standard deviation)
 The Alternative Hypothesis
 H1: S < 2

b) The test a hypothesis about a standard deviation, we use the chi-
 square statistic   
 c^2(pronounced chi-square) = (n-1)s^2/S^2
 where (small case) s is the sample standard deviation
 And so,
 c^2 = (30-1)(1.46)^2/2^2
     = 29(2.1316)/4
     = 15.45

 We use the chi-square distribution with the number of degrees of
 freedom (n-1). The test is one-tailed and to get the critical value
 from the table you need the point with area 0.01 to the left of it.
 From the table, this value is 14.26. And so we see that 15.45 doesnt      
 fall in the rejection zone and we therefore accept the null
 hypothesis.
 Hence, we reject the alternative hypothesis that the standard
 deviation in resolution is significantly below 2 at α = 0.01

Regards.