Question

There are 300 welders employed at Maine Shipyards Corporation. A sample of 30 welders revealed that 18 graduated from a registered welding course. Construct the 95percent confidence interval for the proportion of all welders who graduated from a registered welding course. (Round your answers to 2 decimal places.)

The confidence interval is between and

Answer 1

n=300

p=0.6

μ=np=180

σ=√npq=8.485
the 95 percent confidence interval=180± 1.96*(8.485/√300)=180± 0.96

Answer 2

POPULATION PROPORTION, CONFIDENCE INTERVAL, NORMAL DISTRIBUTION
n: Number of samples = 30
k: Number of Successes = 18
p: Sample Proportion [18/30] = 0.6


significant digits: 1


Confidence Level = 95
"Look-up" Table 'z-critical value' = 2.00
This 'z-critical value' is a Look-up from the Table of Standard Normal Distribution. The Table is organized as a cummulative 'area' from the LEFTcorresponding to the STANDARDIZED VARIABLE z. The Standard Normal Distribution is symmetric (called a 'Bell Curve') which means its an interpretiveprocedure to Look-Up the 'area' from the Table. For the Confidence Level (or Level of Confidence) = 95, there is a LEFT 'area' OUTSIDE. And due to symmetrythere is a RIGHT 'area' OUTSIDE. Using a Look-up from the Table involves adding and subtracting an 'area' which is equal to the Confidence Level. ForSTANDARDIZED VARIABLE z = 2, this corresponds to the LEFT 'area' half of the Confidence Level area = 0.5 * (1 - 95/100) = 0 by a Look-up in the Table forStandard Normal Distribution.
Or alternative; use Excel function: NORMSINV(probability) Returns the inverse of the standard normal cumulative distribution. The distribution has a meanof zero, a standard deviation of one and is symmetrical.

95% Resulting Confidence Interval for 'true mean':
p +/- ('z critical value') * SQRT[p * (1 - p)/n]
= 0.6 +/- 2 * SQRT[0.6 * (1 - 0.6)/30] = [0.4, 0.8]

Answer 3

n: Number of samples = 30
k: Number of Successes = 18
p: Sample Proportion [18/30] = 0.6


significant digits: 1


Confidence Level = 95
"Look-up" Table 'z-critical value' = 2.00
This 'z-critical value' is a Look-up from the Table of Standard Normal Distribution. The Table is organized as a cummulative 'area' from the LEFTcorresponding to the STANDARDIZED VARIABLE z. The Standard Normal Distribution is symmetric (called a 'Bell Curve') which means its an interpretiveprocedure to Look-Up the 'area' from the Table. For the Confidence Level (or Level of Confidence) = 95, there is a LEFT 'area' OUTSIDE. And due to symmetrythere is a RIGHT 'area' OUTSIDE. Using a Look-up from the Table involves adding and subtracting an 'area' which is equal to the Confidence Level. ForSTANDARDIZED VARIABLE z = 2, this corresponds to the LEFT 'area' half of the Confidence Level area = 0.5 * (1 - 95/100) = 0 by a Look-up in the Table forStandard Normal Distribution.


95% Resulting Confidence Interval for 'true mean':
p +/- ('z critical value') * SQRT[p * (1 - p)/n]
= 0.6 +/- 2 * SQRT[0.6 * (1 - 0.6)/30] = [0.4, 0.8]

Answer 4

POPULATION PROPORTION, CONFIDENCE INTERVAL, NORMAL DISTRIBUTION
n: Number of samples = 30
k: Number of Successes = 18
p: Sample Proportion [18/30] = 0.6


significant digits: 1


Confidence Level = 95
"Look-up" Table 'z-critical value' = 2.00
This 'z-critical value' is a Look-up from the Table of Standard Normal Distribution. The Table is organized as a cummulative 'area' from the LEFTcorresponding to the STANDARDIZED VARIABLE z. The Standard Normal Distribution is symmetric (called a 'Bell Curve') which means its an interpretiveprocedure to Look-Up the 'area' from the Table. For the Confidence Level (or Level of Confidence) = 95, there is a LEFT 'area' OUTSIDE. And due to symmetrythere is a RIGHT 'area' OUTSIDE. Using a Look-up from the Table involves adding and subtracting an 'area' which is equal to the Confidence Level. ForSTANDARDIZED VARIABLE z = 2, this corresponds to the LEFT 'area' half of the Confidence Level area = 0.5 * (1 - 95/100) = 0 by a Look-up in the Table forStandard Normal Distribution.
Or alternative; use Excel function: NORMSINV(probability) Returns the inverse of the standard normal cumulative distribution. The distribution has a meanof zero, a standard deviation of one and is symmetrical.

95% Resulting Confidence Interval for 'true mean':
p +/- ('z critical value') * SQRT[p * (1 - p)/n]
= 0.6 +/- 2 * SQRT[0.6 * (1 - 0.6)/30] = [0.4, 0.8]