The data below gives the mean price (in cents) of a liter of regular gasoline at self-service filling stations at a sample of six urban centers in Canada in May 2011. Urban area Price/liter
St. John, NL 137.1
Halifax 133.6
Montreal 136.2
Regina 132.3
Saskatoon 132.2
Yellowknife 138.9

Required:
a. Find the sample mean (in cents).
b. Find the sample standard deviation (in cents).
c. Find a 90% confidence interval for the mean gasoline price per liter across urban Canada in May 2011. Provide the lower and upper limits for your confidence interval.
d. Find a 95% confidence interval for the mean gasoline price per liter across urban Canada in May 2011. Provide the lower and upper limits for your confidence interval.

135.05 ; 2.762 ; (133.201, 136.899) ; (132.840, 137.260)

Step-by-step explanation:

Given the data:

Urban area Price/liter

St. John, NL 137.1

Halifax 133.6

Montreal 136.2

Regina 132.3

Saskatoon 132.2

Yellowknife 138.9

Mean = (137.1 + 133.6 + 136.2 + 132.3 + 132.2 + 138.9) / 6

= 810.3 / 6

= 135.05

Standard deviation (σ) : √Σ(X - mean)²/n-1

(137.1 - 135.05)^2 + (133.6 - 135.05)^2 + (136.2 - 135.05)^2 + (132.3 - 135.05)^2 + (132.2 - 135.05)^2+ (138.9 - 135.05)^2 = 38.135

√(38.135 / (6 - 1))

= √7.627

= 2.7617023

= 2.762

C.)

90% confidence interval of the mean :

Mean ± Zcrit * σ/√n

Zcrit at 90% = 1.64

135.05 ± 1.64 * (2.762/√6)

135.05 ± 1.64 * 1.1275817

Lower Value = (135.05 - (1.64 * 1.1275817)) = 133.201

Upper Value = (135.05 + (1.64 * 1.1275817))) = 136.899

Confidence interval = (133.201, 136.899)

D.)

95% confidence interval of the mean :

Mean ± Zcrit * σ/√n

Zcrit at 95% = 1.96

135.05 ± 1.96 * (2.762/6)

135.05 ± 1.96 * 1.1275817

Lower Value = (135.05 - (1.96 * 1.1275817)) = 132.840

Upper Value = (135.05 + (1.96 * 1.1275817)) = 137.260

Confidence interval = (132.840, 137.260)

i believe the answer is c.

a28=58

step-by-step explanation:

a1=a1+(n-1)d

a28= 4+ (28-1)2

a28=4+(27)(2)

a28=4+54

a28=58