Question

3. The mean age of sample of 100 cars from a certain manufacturer is found to be eleven years. If the sample standard deviation of car ages is 5 years, give a 99% confidence interval for the mean age of cars from this manufacturer. 4. The sample standard deviation of the ages of a random sample of 40 television sets in a neighborhood is 3 years. Find a 95% confidence interval for the standard deviation of the entire population of televisions in this neighborhood. Assume that these ages are randomly distributed.

Answer 1

3)

sample std dev ,    s =    5.0000
Sample Size ,   n =    100
Sample Mean,    x̅ =   11.0000

Level of Significance , α = 0.01

degree of freedom= DF=n-1= 99

't value=' tα/2= 2.626 [Excel formula =t.inv(α/2,df) ]

Standard Error , SE = s/√n = 5/√100= 0.5000

margin of error , E=t*SE = 2.6264 * 0.5000 = 1.313

confidence interval is

Interval Lower Limit = x̅ - E = 11.00 - 1.3132 = 9.6868

Interval Upper Limit = x̅ + E = 11.00 - 1.3132 = 12.3132

99% confidence interval is ( 9.69 < µ < 12.31 )

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4)

Sample Size,   n=   40
Sample Standard Deviation,   s=   3.0000
Confidence Level,   CL=   0.95
      
      
Degrees of Freedom,   DF=n-1 =    39
alpha,   α=1-CL=   0.05
alpha/2 ,   α/2=   0.025
Lower Chi-Square Value=   χ²1-α/2 =   23.654
Upper Chi-Square Value=   χ²α/2 =   58.120
      

confidence interval for std dev is       
lower bound= √[(n-1)s²/χ²α/2] =   √(39*3² / 58.1201)=   2.4575
      
      
upper bound= √[(n-1)s²/χ²1-α/2] =   √(39*3² / 23.6543)=   3.8521
      

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