In: Statistics and Probability
Given the Data set: 8.5, 8.7, 8.9, 8.2, 8.3, 8.4, 8.5 – Use this set for parts a to d. This is to be considered as set2 for part c
[Find the Mean, Mode, Median, Range, Midrange and Standard Deviation for the Data Set
x | x^2 | |
1 | 8.2 | 67.24 |
2 | 8.3 | 68.89 |
3 | 8.4 | 70.56 |
4 | 8.5 | 72.25 |
5 | 8.5 | 72.25 |
6 | 8.7 | 75.69 |
7 | 8.9 | 79.21 |
total | 59.5 | 506.09 |
Mean | 8.5 | |
SD | 0.238 | |
Mode | 8.5 | value with highest freq |
Median | 8.5 | (N+1)/2 th value |
Range | 0.7 | Max - min |
Midrange | 8.55 | (min + max)/2 |
Lower Quartile(Q1) | 8.3 | (N+1)/4 th value |
Upper quartile (Q3) | 8.7 | 3(N+1)/2 th value |
IQR | 0.4 | Q3 - Q1 |
Mean =
SD =
Determine if the following data set has an outlier that can be disregarded
Outlier is when a value lies outside the range (Q1 - 1.5IQR, Q3 + 1.5IQR)
Q1 - 1.5IQR = 7.7
Q3 + 1.5IQR = 9.3
since no values outside this range there are no outliers.
Using the given data set#2 from the given data and set #1 being (?⃐ = 8.7 , S = 0.47 and N = 7) determine if the two data sets are statistically the same or different using an 90% confidence interval.
Assuming the data comes from a normal population
We have
Null hyptohesis: The means are same for both sets
Alternative: The means are different
1 | 2 | |
n | 7 | 7 |
mean | 8.7 | 8.5 |
SD | 0.47 | 0.238 |
Var | 0.221 | 0.057 |
Pooled variance =
= 0.13878
Test Stat =
Level of significance (1 - 90% = 10%
Critical value at 10% =
= 1.7823 ......Using t-dist with df = 12 and p =0.025
then we substitute all the values
(-0.1549,0.5549)
Here since the null difference = 0 lies within the CI
We do not reject the null hypothesis thus concluding that at 10% level the means are significantly same.
Find the range for the sample mean value with 80% confidence interval. Make sure µ is isolated
Here since we do not have the population SD we will use a t-dist
80% confidence interval for population mean
Where the critical value at 20% (1-0.8)=
= 1.4398
Substituting the rest of values