10 randomly selected automotive batteries is shown to the right. 1.77 1.81 1.52

Question

10

randomly selected automotive batteries is shown to the right.

1.77

1.81

1.52

1.67

1.72

1.93

1.38

1.58

1.41

2.06

Assume the sample is taken from a normally distributed population. Construct

90

%

confidence intervals for (a) the population variance

sigma

squaredand (b) the population standard deviation

sigma

.

(a) The confidence interval for the population variance is

(nothing

,nothing

).

(Round to three decimal places as needed.)

Interpret the results. Select the correct choice below and fill in the answer box(es) to complete your choice.

(Round to three decimal places as needed.)

A.

With

10

%

confidence, it can be said that the population variance is between

nothing

and

nothing

.

B.

With

90

%

confidence, it can be said that the population variance is less than

nothing

.

C.

With

90

%

confidence, it can be said that the population variance is between

nothing

and

nothing

.

D.

With

10

%

confidence, it can be said that the population variance is greater than

nothing

.

(b) The confidence interval for the population standard deviation is

(nothing

,nothing

).

(Round to three decimal places as needed.)

Interpret the results. Select the correct choice below and fill in the answer box(es) to complete your choice.

(Round to three decimal places as needed.)

A.

With

90

%

confidence, you can say that the population standard deviation is between

nothing

and

nothing

hours of reserve capacity.

B.

With

10

%

confidence, you can say that the population standard deviation is greater than

nothing

hours of reserve capacity.

C.

With

10

%

confidence, you can say that the population standard deviation is between

nothing

and

nothing

hours of reserve capacity.

D.

With

90

%

confidence, you can say that the population standard deviation is less than

nothing

hours of reserve capacity

Explanation / Answer

Part a

Here, we have to find 90% confidence interval for the population variance. Formula for confidence interval is given as below:

(n – 1)*S2 / 2/2, n – 1 < 2 < (n – 1)*S2 / 21 - /2, n – 1

From given data, we have

Sample size = n = 10

Degrees of freedom = n – 1 = 10 – 1 = 9

Sample standard deviation = S = 0.21905098

Confidence level = 90%, c = 0.90, = 1 – c = 1 – 0.90 = 0.10, /2 = 0.10/2 = 0.05

(n – 1)*S2 = 9*0.21905098^2 = 0.43185

21 - /2, n – 1 = 3.3251 (by using Chi square table or excel)

2/2, n – 1 = 16.9190 (by using Chi square table or excel)

(n – 1)*S2 / 2/2, n – 1 < 2 < (n – 1)*S2 / 21 - /2, n – 1

0.43185 / 16.9190 < 2 < 3.3251/3.3251

0.0255 < 2 < 0.1299

The confidence interval for the population variance is (0.026, 0.130).

Interpretation:

With 90% confidence, it can be said that the population variance is between (0.026, 0.130).

Part b

Confidence interval for population standard deviation is given as below:

Sqrt[(n – 1)*S2 / 2/2, n – 1 ] < 2 < sqrt[(n – 1)*S2 / 21 - /2, n – 1 ]

Sqrt(0.0255) < < sqrt(0.1299)

0.1598 < < 0.3604

The confidence interval for the population standard deviation is (0.1598, 0.3604).

Interpretation:

With 90% confidence, it can be said that the population standard deviation is between (0.1598, 0.3604) hours of reserve capacity.

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