Math: Confidence Intervals, someone please hellpp: Construct a 95% confidence interval for n = 100, p = 0.8 ..?

Question

1.) Construct a 95% confidence interval for μ = 120, σ = 15
	
-100.1 to 151.2
-90.6 to 149.4
-99.8 to 149.4
-110.3 to 156.3

2.) Construct a 95% confidence interval for n = 100, p = 0.8
	
-85.4 to 120.6
-67.13 to 99.8
-70.42 to 94.85
-72.16 to 87.84

3.) In a survey of a random sample of 1000 teenagers in Nanaimo, B.C., it was found that 125 of these teenagers had never travelled outside of their own province. Construct a 95% confidence interval for this data.
	
-111.2 to 137.2
-104.502 to 145.498
-87.502 to 145.534
-98.487 to 151.673

Sanjay“Natsu”Jagarwal · Accepted Answer

1.  Mean  μ = 120  ,  standard deviation σ = 15 , condition 95% confidence interval .

sample statistic =  we choose the mean weight in our sample (120) as the sample statistic.

Compute alpha (α): α = 1 - (confidence level / 100) =1 - 95/100 = 0.05

the critical probability (p*): p* = 1 - α/2 = 1 - 0.05/2 = 0.975

The value of the random variable Y is:
            Y = { 1/[ σ * sqrt(2π) ] } * e^{-(x - μ)^2/2σ^2}
  using this , upper bond is 149.399  ,  p  =0.975
              lower bond  90.601  , p = 1-p = 0.025

hence answer is -90.6 to 149.4.

2. 95% confidence interval for n = 100, p = 0.8

degrees of freedom (df): df = n - 1 = 100 - 1 = 99

standard deviation ,  σ =√[ P(1 - P) / n ]  =  √(0.8)(1-0.8)/n = 0.04

Compute alpha (α): α = 1 - (confidence level / 100) = 1 - 0.90 = 0.10

Critical value t = 0.845

standard error (SE) of the mean is: SE = σ / sqrt( n ) = 0.04 / sqrt(100) = 0.004

margin of error (ME): ME = Critical value x Standard error = 0.845x0.004 = 0.00338

3. n = 1000 , p = 125/1000 = 0.125 ,

df = 1000 -1 = 999

standard deviation ,  σ =√[ P(1 - P) / n ]  =  √(0.125)(1-0.125)/n = 0.01045

Compute alpha (α): α = 1 - (confidence level / 100) =1 - 95/100 = 0.05

standard error =  0.01045/√1000 = 0.0003307

Critical value t  = -1.151

margin of error (ME): ME = Critical value x Standard error = 3.806357x10^(-4)

paramvenu · Answer

2nd and 3rd questions are just now only answered.

1) 95% confidence interval for population mean is
Sample mean +/- z-score for 95% confidence*Standard error of mean
Sample mean = 120 (The given mean is NOT Mu but it is xbar)
z-score for 95% confidence = 1.96
Standard error of mean = Standard deviation/sqrt sample size 
Since the sample size is not given the given standard deviation is assumed as standard error of mean = 15
Therefore the confidence interval is
120 +/- 1.96*15
120 +/- 29.4
lower limit is 120-29.4 = 90.6
upper limit is 120+29.4 = 149.4
The confidence interval is 90.6 to 149.4
<<<<<<<<<<<<<<<
If the sample size is known, calculate the standard error of mean  as stated above and substitute in the above formula

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Math: Confidence Intervals, someone please hellpp: Construct a 95% confidence interval for n = 100, p = 0.8 ..?

2 Answers