Mean Standard Deviation N Lower class 11.61 2.67 196 Working class 12.80 2.85 62
ID: 3066474 • Letter: M
Question
Mean
Standard Deviation
N
Lower class
11.61
2.67
196
Working class
12.80
2.85
626
Middle class
14.45
3.08
484
Upper class
15.45
4.05
60
The 2010 GSS provides these statistics above for the average years of education for lower-, working-, middle-, and upper-class respondents and their associated standard deviations.
1. Construct the 90% confidence interval for the mean number of years of education for lower-class and middle-class respondents (10 pts)?
2. Construct the 99% confidence interval for the mean number of years of education for lower-class and middle-class respondents (10 pts)?
3. As our confidence in the result increases, how does the size of the confidence interval change? Explain why this is so (10 pts).
Mean
Standard Deviation
N
Lower class
11.61
2.67
196
Working class
12.80
2.85
626
Middle class
14.45
3.08
484
Upper class
15.45
4.05
60
Explanation / Answer
1)
Step 1: Find /2
Level of Confidence = 90%
= 100% - (Level of Confidence) = 10%
/2 = 5% = 0.05
Step 2: Find degrees of freedom and t/2
Degrees of freedom = smaller of (n1 - 1 , n2 - 1 ) = smaller of (195 , 483) = 195
Calculate t/2 by using t-distribution with degrees of freedom (DF) = 195 and /2 = 0.05 as right-tailed area and left-tailed area.
Step 3: Calculate Confidence Interval
t/2 = 1.65269
Standard Error = (s)²/n + (s)²/n = 0.05597 = 0.2365
Lower Bound = (x - x) - t/2•( (s)²/n + (s)²/n ) = (11.61 - 14.45) - (1.65269)(0.2365) = -3.2309
Upper Bound = (x + x) + t/2•( (s)²/n + (s)²/n ) = (11.61 - 14.45) + (1.65269)(0.2365) = -2.4490
Confidence Interval = (-3.2309, -2.4490)
2)
Step 1: Find /2
Level of Confidence = 99%
= 100% - (Level of Confidence) = 1%
/2 = 0.5% = 0.005
Step 2: Find degrees of freedom and t/2
Degrees of freedom = smaller of (n1 - 1 , n2 - 1 ) = smaller of (195 , 483) = 195
Calculate t/2 by using t-distribution with degrees of freedom (DF) = 195 and /2 = 0.005 as right-tailed area and left-tailed area.
Step 3: Calculate Confidence Interval
t/2 = 2.60113
Standard Error = (s)²/n + (s)²/n = 0.05597 = 0.2365
Lower Bound = (x - x) - t/2•( (s)²/n + (s)²/n ) = (11.61 - 14.45) - (2.60113)(0.2365) = -3.4553
Upper Bound = (x + x) + t/2•( (s)²/n + (s)²/n ) = (11.61 - 14.45) + (2.60113)(0.2365) = -2.2246
Confidence Interval = (-3.4553, -2.2246).
3) As we increase our confidence level , then we increase the size of the range about our estimate