Question 555:
1Answer:
No answer provided yet.Part a. 2-Sample z
30 IQ scores from two classes which have normal distribution were samples to see if there was a difference in average IQ. Below are the IQ scores.
Class A |
Class B |
100 |
70 |
88 |
88 |
75 |
75 |
115 |
115 |
120 |
120 |
112 |
108 |
108 |
75 |
120 |
85 |
115 |
100 |
122 |
102 |
95 |
95 |
99 |
80 |
101 |
101 |
105 |
105 |
110 |
110 |
117 |
100 |
79 |
79 |
102 |
102 |
108 |
108 |
98 |
98 |
125 |
99 |
100 |
98 |
93 |
93 |
105 |
105 |
103 |
103 |
115 |
115 |
130 |
103 |
91 |
91 |
90 |
90 |
95 |
90 |
The descriptive statistics for the above IQ scores are for Class A Mean: 104.53, SD 13.24 and for Class B Mean : 96.77 and SD 12.55.
- First we generate a pooled standard deviation of the difference using the variance (squaring the standard deviation) since we're assuming there is a difference between variances.
- The variances are 175.43 and 157.43
- In assuming unequal variances we use the formula for the pooled standard deviation
- SQRT( (var1/n1) + (var2/n2) )
- SQRT( (175.43 /30) + (157.43/30) ) = 3.331 which is the pooled standard deviation.
- The difference between the means is 104.53-13.24 = 7.77 making the test statistic = 7.77/3.331= 2.66. Since we're using the z-test, this is our z-score.
- We can use a z-table or the z-score to percentile calculator and enter 2-sided. We get a percent of area of .7814 which is a p-value of .007814 or well less than .05.
The p-value of .0078 is below our rejection criteria of .05 meaning we would reject the null hypothesis that there is no difference between Class IQ's. This means we would say there is a significant difference in IQ scores between the two classes.
Part b. 2-Sample t-Test
The lunch-room lady wanted to know if two lunch periods had more food on their lunch-trays. The weight in pounds of two samples of one from the early lunch and one from the later lunch were sampled and are below.
Early |
Late |
1.65 |
1.73 |
0.75 |
1.06 |
1.03 |
3.01 |
1.12 |
1.4 |
1.05 |
1.15 |
1.02 |
1.13 |
1.25 |
1.41 |
1.26 |
1.73 |
1.01 |
1.63 |
1.03 |
1.56 |
The descriptive statistics for the above weights above are for Early Mean: 1.117lbs, SD .235 and for the Late Lunch Mean : 1.581lbs and SD .559.
- First we generate a pooled standard deviation of the difference using the variance (squaring the standard deviation) since we're assuming there is a difference between variances.
- The variances are .05527 and .3124
- In assuming unequal variances we use the formula for the pooled standard deviation
- SQRT( (var1/n1) + (var2/n2) )
- SQRT( (.05527 /10) + (.3124/10) ) = .1918 which is the pooled standard deviation of the difference.
- There are 10+10-2 = 18 degrees of freedom.
- The difference between the means is 1.581-1.117 = 4.64 making the test statistic = 4.64/.1918= 2.419. Since we're using the t-test, this is our test statistic t. We can use the percentile from t-score calculator here http://www.usablestats.com/calcs/tdist and we get a p-value of .0264.
- The p-value of .0264 is below our rejection criteria of .05 meaning we would reject the null hypothesis that there is no difference between lunch weights. This means we would say there is a significant difference in lunch weights between the two lunch times.
We can also check the results using the 2-sample t-test calculator here
http://www.usablestats.com/calcs/2samplet
Part C Two Proportions. Variables
In a random sample of 200 people from
The samples are large enough that we can use the Normal Approximation to the Binomial to answer the difference between two proportions. We see that 60% in
You divide this difference by the square-root of a denominator which accounts for the chance which will provide you with a z-score. The denominator is the square root of:
1 1
-- + -- * PQ
n_{1} n_{2}
Where P = (x1 + x2)/(n1+n2) Q is 1-P. The x's are just the number of people who favor winter in each city and the n's are the sample sizes.
P = (120+240)/(200+500) = .514
Q = 1-.514 = .486
PQ = .486*.513 = .2498
1/n1 + 1/n2 = 1/200 + 1/500 = .007
So multiply .2498 * .007= 0.00175
Now the square root of this is SQRT(0.00175) = .04183
So the equation is the observed difference .12/.04183 = 2.87
That result is the z-score which is your test-statistic. You now look this value up using the z-score to percentile calculator using the 2-sided area. You should get 0.413% or a p-value of .00413, meaning there is about a .04% chance the difference is due to chance. In other words, we would conclude that more