Inference for Variance

Inference for One and Two Variances

Name(s):
Name 1, Name 2, Name 3, Name 4

Questions
Inference for One and Two Variances

The standard deviation of egg weights for young hens in the population is 5.3 grams (based on a database maintained by an egg producers’ trade-group) and the standard deviation of older hens in the population is 6.4 grams (again, from the database). An egg producer who uses free-range and organic methods seeks to understand the variability of the eggs produced in such a manner (realizing that this method of egg production may well have a different variability than standard production). The rationale of this is because of the labeling mandates of the way in which “small”, “medium,” and “large” egg labels can be used (and if young hens are cost effective).
1. The egg producer collected data from 101 young hens and found that the standard deviation was 6.2 grams. State the null and alternative hypotheses to test if the variance of the eggs of young hens is equal to the variance of young hens of the population in general. Answer:
```
 Ho = young = young pop
 Ha =  young <> young in pop
```
  1. What is/are the critical value(s) for the test of the null hypothesis at a .05 level of significance?
```
qchisq(c(.025,.975),101-1)
```
```
## [1]  74.22193 129.56120
```
  2. What is the value of the test statistic for the test of the null hypothesis?
```
           100*(6.2^2/5.3^2)
```
```
## [1] 136.8459
```
  3. Is there a statistical difference between the variances? Explain.
    Answer:
```
     I can conclude that there is  a significant difference of the weight of the eggs produced by the sample of the hen population with 95% confidence because of our P-Value of .017. Therefore  we reject the null.
```
  4. What is the 95% confidence interval for the population variance of this particular egg producer’s young hens?
```
c((((101-1)*6.2^2)/129.56),(((101-1)*6.2^2)/74.22))
```
```
## [1] 29.66965 51.79197
```
2. The egg producer also has 61 older hens whose eggs have a standard deviation of 5.8 grams. State the null and alternative hypotheses to test if the variance of his older hens’ eggs is less than the variance of older hens of the population in general.
  Answer:
```
  Ho = Older pop > older sample 
  Ha = Older pop < older sample 
```
  1. What is/are the critical value(s) for the test of the null hypothesis at a .05 level of significance?
```
 c(0,qchisq(.05,61-1))
```
```
## [1]  0.00000 43.18796
```
  2. What is the value of the test statistic for the test of the null hypothesis?
```
(61-1)*(5.8^2/6.4^2)
```
```
## [1] 49.27734
```
  3. Is the variance of this egg producer’s older hens’ eggs statistically less than the variance of the population in general? Explain.
    0.1629907 Fail to reject
  4. What is the 95% confidence interval for the population variance of this particular egg producer’s older hens?
```
c(0,((61-1)*5.8^2)/43.18796)
```
```
## [1]  0.00000 46.73525
```
3. Now, the egg producer is interested in comparing variances in egg weights from older and younger hens for the free-range style of farming. State the null and alternative hypotheses to test the equality of population variances (for the free-range hens).
  Answer:
```
  Ho = Older = Younger
  Ha = Older <> Younger
```
  1. What is the value of the test statistic for the test of the null hypothesis?
```
(6.2^2)/(5.8^2)
```
```
## [1] 1.142687
```
  2. What is the critical value for \(\alpha\) = .05?
```
qf(c(.025,.975), 100, 60, lower.tail=TRUE)
```
```
## [1] 0.642043 1.599037
```
  3. What is the conclusion? Explain.
    Answer:
Talking Internet

35.8 14.0

22.2 12.5

4.0 16.4

32.6 1.9

8.5 9.9

42.5 5.4

8.0 1.0

3.8 15.2

30.0 4.0

12.8 4.7

10.3

35.5
Many smartphones have earned a bad reputation for exceptionally poor battery life. Battery life between charges for the Motorola Droid Razr Max averages 20 hours when the primary use is talk time and 7 hours when the primary use is Internet applications (The Wall Street Journal). Since the mean hours for talk time usage is greater than the mean hours for Internet usage, the question was raised as to whether the variance in hours of usage is also greater when the primary use is talk time. Sample data showing battery hours of use for the two applications is above.
1. Formulate hypotheses about the two population variances that can be used to determine if the population variance in battery hours of use is greater for the talk time application.
  Answer:
```
   Ho = variance of usage for talking is not greater than the variance of usage for internet
   Ha = Variance of usage for talking is greater than the variance of usage for internet
```
2. What are the standard deviations of battery hours of use for the two samples?
```
       talking <- c(35.8,22.2,4,32.6,8.5,42.5,8.0,3.8,30.0,12.8,10.3,35.5)
       internet <- c(14.0,12.5,16.4,1.9,9.9,5.4,1.0,15.2,4.0,4.7)
       sd(talking)
```
```
## [1] 14.13365
```
```
       sd(internet)
```
```
## [1] 5.769845
```
3. Conduct the hypothesis test for the variance from part a: include the value of the test statistic and the \(p\)-value.
```
test <- var.test(x=talking, y=internet, alternative = "greater")
test
```
```
## 
##  F test to compare two variances
## 
## data:  talking and internet
## F = 6.0004, num df = 11, denom df = 9, p-value = 0.005994
## alternative hypothesis: true ratio of variances is greater than 1
## 95 percent confidence interval:
##  1.934062      Inf
## sample estimates:
## ratio of variances 
##           6.000401
```
4. Using a .05 level of significance, what is your conclusion? Explain.
  Answer: 0.0059939 Reject the Null Hypothesis

Talking	Internet
35.8	14.0
22.2	12.5
4.0	16.4
32.6	1.9
8.5	9.9
42.5	5.4
8.0	1.0
3.8	15.2
30.0	4.0
12.8	4.7
10.3
35.5

Inference for Variance

Ken Kelley’s Statistics for Managerial Decision Making