Answer to Question #196254 in Statistics and Probability for Leo

Question

 A sample of 60 Grade 9 students ages was obtained to estimate the mean age of all Grade 9 students. x̄ = 1.5 years and the population standard deviation is 4.

i. What is the point estimate for u?

ii. Find the 95% confidence interval for u.

iii. What conclusion can you make based on each estimate?

Answer 1

part (i)

point estimate:-

It provides single value that is used to approximate an unknown population parameters

(a) same mean (x̄) = population mean (μ) = 1.5

(b) sample proportion(P) = population proportion(p)

(c)sample standard deviation(s)= population standard deviation(σ)

σ = s.[√n/√(n-1)]

=4[√60/√(60-1)]

=4.03375587272

(d)sample varience (s²)= population varience(σ²)

σ^{2 =} 16.2711864407.

=================================================================

part(ii)

Confidence Interval

formula:-

CI =X̄ ± Z×(s/√n)

   =1.5 ± 1.9600×(4/√60)

   =1.5 ± 1.012

 simplify for + and - 

we get

1.5 - 1.012=0.488

and

1.5 + 1.012=2.512

so Confidence Interval

[0.488 – 2.512]...............answer

==========================================================

The margin of error depends on the size of the sample used to construct the confidence interval, whether the population standard deviation is known, and the level of confidence chosen.

The resulting interval is known as a confidence interval. A confidence interval is constructed with a specified level of probability. For example, suppose you draw a sample of stocks from a portfolio, and you construct a 95 percent confidence interval for the mean return of the stocks in the entire portfolio:

(lower limit, upper limit) = (0.488 , 2.512)