Question

The proper operation of typical home appliances requires voltage levels that do not vary much. Listed below are 9 voltage levels (in volts) at a random house on 9 different days.

120.3 119.5 120 120 120 120.1 120 120 119.8

The mean of the data set is 120; the sample standard deviation is 0.22; if the normality plot is not provided you may assume that the voltages are normally distributed.

1. Construct a 99% confidence interval for the variance (standard deviation) of all voltages in the house.
   1. Procedure: Select an answer: One mean Z procedure, One variance χ² procedure, One mean T procedure, One proportion Z procedure

      Hint: In the question, look for the keywords such as mean/average, proportion/percentage, variance/standard deviation!

   2. Assumptions: (select everything that applies)
      • Sample size is greater than 30
      • Population standard deviation is known
      • Population standard deviation is unknown
      • The number of positive and negative responses are both greater than 10
      • Normal population
      • Simple random sample
   3. 1. Unknown parameter: Select an answerμ, population meanp, population proportionσ², population variance
      2. Point estimate: Select an answersample proportion, p̂sample mean, x̄sample variance, s² = (Round the answer to 4 decimal places)
      3. Confidence level % → α=α= → α2=α2= and 1−α2=1-α2=
         Critical values: right= left= (Round the answer to 3 decimal places)
      4. Margin of error (if applicable): (Round the answer to 4 decimal places)
      5. Lower bound: (Round the answer to 4 decimal places)
      6. Upper bound: (Round the answer to 4 decimal places)
      7. Confidence interval:(, )
      8. Interpretation: We are % confident that the true population parameter is between and .

   4. Based on the confidence interval, is it reasonable to believe that the population variance is less than 0.01? Explain.

   5. Select an answer: Yes or No , because Select an answer: the entire interval is below, the entire interval is above, a part or the entire interval is below, a part or the entire interval is above 0.01.

Answer 1

GIVEN MEAN OF THE DATA SET IS 120 AND STANDARD DEVIATION OF THE DATA SET IS 0.22

a) i) One variance χ² procedure.

Reason: Because we use  χ² to calculate one population variance interval.

ii) Population standard deviation is unknown,Normal population and Simple random sample.

iii) σ², population variance

iv) Point estimate: s^2= 0.0484

Confidence level :

left end confidence interval value: αleft end = α/2

αleft end = 0.01/2

αleft end = 0.005

αright end = 1 - α/2

α right end = 1 - 0.01/2

α right end = 0.995

LEFT END critical χ2 value for 0.005

χ2 (0.005) = 21.9549

Value can be found on Excel using =CHIINV(0.005,8)

RIGHT end critical χ2 value for 0.995 χ2(0.995) = 1.3444

Value can be found on Excel using =CHIINV(0.995,8)

iv) MARGIN OF ERROR IS NOT APPLICABLE HERE

v) Lower bound  = ((n - 1)s^2/χ^2α/2)

Lower bound = (8)*(0.0484)/21.9549)

= 0.3872/21.9549

= 0.0176

Lower bound = 0.0176

vi) Upper bound =((n - 1)s2/χ21 - α/2)

Upper bound = (8)(0.0484)/1.3444)

Upper bound= 0.3872/1.3444

Upper bound = 0.2880095209759

Upper bound= 0.2880

vii) Confidence interval is ( 0.0176 , 0.2880)

viii) We are 99% confident that the true population parameter is between 0.0176 and 0.2880

YES , because the entire interval is above 0.01