Question

1.  Find the margin of error for the mean weight of medium Antonio’s cheese pizza if a sample of size n = 16 produces a sample standard deviation of 4.220 g, assuming 95% confidence. Round to two decimals.

2. The mean and standard deviation for voltages of power packs labeled as 12 volts for a sample of 25 are as follows:

	=	12.34
s	=	0.3

Please develop a 95.00% confidence interval for the sample above. (Round to 3 decimal places.)
Lower limit = ?

Upper limit = ?

3. The actual voltages of power packs labeled as 12 volts for a sample are as follows: 11.90, 11.79, 11.38, 11.79, 12.21, 11.95, 11.38, 11.01.
Please develop a 90.00% confidence interval for the sample above.

Lower limit = ?

Upper limit = ?

4. Assuming the random variable X is normally distributed, compute the lower and upper limits of the 90% confidence interval for the population mean if a random sample of size n=7 produces a sample mean of 12 and sample standard deviation of 6.00. Round to two decimals.

Lower Limit = ?,

Upper Limit = ?

5. A random sample of 22 scalpers’ ticket prices for a rock concert has sample mean $44.00 and sample standard deviation of $6.02. What is the upper and lower limit of the 95% confidence interval of mean scalpers’ ticket prices? Round to two decimals.

Lower Limit = ?

Upper Limit = ?

Answer 1

1. n = 16

sample sd = s = 4.220

95 % CI

df = 15

t value = TINV(0.05,15) = 2.131

ANS: Margin of error = 2.25

2. n = 25, mean = 12.34, sd = 0.3

t value = TINV(0.05,24) = 2.064

CI = mean +/-E = 12.34 +/- 0.124 = (12.216 , 12.464)

3. n = 8,

mean = sum of all terms / no of terms = 93.41/ 8 = 11.6763

sample sd = s

data	data-mean	(data - mean)2
11.90	0.2237	0.05004169
11.79	0.1137	0.01292769
11.38	-0.2963	0.087793689999999
11.79	0.1137	0.01292769
12.21	0.5337	0.28483569
11.95	0.2737	0.07491169
11.38	-0.2963	0.087793689999999
11.01	-0.6663	0.44395569

t value at 90% = TINV(0.10,7) = 1.895

CI = mean +/-E = 11.6763 +/- 0.260 = (11.416 , 11.936)

4.  n = 7, mean = 12, sd = 6

t value = TINV(0.1,6) = 1.943

CI = mean +/-E = 12 +/- 4.407 = (7.593 , 16.407)

5.

n = 22, mean = 44, sd = 6.02

t value = TINV(0.1,21) = 2.080

CI = mean +/-E = 44 +/- 2.669 = (41.331 , 46.669)