Question

The method of tree ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population of x values has an approximately normal distribution. 1285 1250 1187 1236 1268 1316 1275 1317 1275

(a) Use a calculator with mean and standard deviation keys to find the sample mean year x and sample standard deviation s. (Round your answers to the nearest whole number.)

x = ____A.D.

s = _____yr

(b) Find a 90% confidence interval for the mean of all tree ring dates from this archaeological site. (Round your answers to the nearest whole number.)

lower limit A.D.

upper limit A.D.

________________________________________________________________________

How much does a sleeping bag cost? Let's say you want a sleeping bag that should keep you warm in temperatures from 20°F to 45°F. A random sample of prices ($) for sleeping bags in this temperature range is given below. Assume that the population of x values has an approximately normal distribution.

50	90	95	110	85	60	30	23	100	110
105	95	105	60	110	120	95	90	60	70

(a) Use a calculator with mean and sample standard deviation keys to find the sample mean price x and sample standard deviation s. (Round your answers to two decimal places.)

x =	$
s =	$

(b) Using the given data as representative of the population of prices of all summer sleeping bags, find a 90% confidence interval for the mean price μ of all summer sleeping bags. (Round your answers to two decimal places.)

lower limit	$
upper limit	$

Answer 1

Question 1

	Values ( X )	Σ ( Xi- X̅ )2
	1285	300.4433
	1250	312.1123
	1187	6507.1165
	1236	1002.7799
	1268	0.1111
	1316	2336.1079
	1275	53.7773
	1317	2433.7745
	1275	53.7773
Total	11409	13000.0001

Part a)

Mean X̅ = Σ Xi / n
X̅ = 11409 / 9 = 1267.6667 1268

Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 13000.0001 / 9 -1 ) = 40.3113 40

Part b)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.1 /2, 9- 1 ) = 1.86
1267.6667 ± t(0.1/2, 9 -1) * 40.3113/√(9)
Lower Limit = 1267.6667 - t(0.1/2, 9 -1) 40.3113/√(9)
Lower Limit = 1242.6798   1243
Upper Limit = 1267.6667 + t(0.1/2, 9 -1) 40.3113/√(9)
Upper Limit = 1292.6536   1293
90% Confidence interval is ( 1243 , 1293 )

Question 2

	Values ( X )	Σ ( Xi- X̅ )2
	50	1098.9225
	90	46.9225
	95	140.4225
	110	720.9225
	85	3.4225
	60	535.9225
	30	2824.9225
	23	3618.0225
	100	283.9225
	110	720.9225
	105	477.4225
	95	140.4225
	105.0	477.4225
	60	535.9225
	110	720.9225
	120	1357.9225
	95	140.4225
	90	46.9225
	60	535.9225
	70	172.9225
Total	1663	14600.55

Part a)

Mean X̅ = Σ Xi / n
X̅ = 1663 / 20 = 83.15

Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 14600.55 / 20 -1 ) = 27.72

Part b)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.1 /2, 20- 1 ) = 1.729
83.15 ± t(0.1/2, 20 -1) * 27.7209/√(20)
Lower Limit = 83.15 - t(0.1/2, 20 -1) 27.7209/√(20)
Lower Limit = 72.43
Upper Limit = 83.15 + t(0.1/2, 20 -1) 27.7209/√(20)
Upper Limit = 93.87
90% Confidence interval is ( 72.43 , 93.87 )