Subjective Test Questions Student Name Institutional Affiliation Q1 No, it is not reasonable to calculate a confidence interval for the data. Reason: the popula
Running head: SUBJECTIVE TEST QUESTIONS 0
Subjective Test Questions
Student Name
Institutional Affiliation
Q1
No, it is not reasonable to calculate a confidence interval for the data.
Reason: the population standard deviation is not given.
Q2
We are given:
Margin of estimate, e = 0.10(100) hours
Level of confidence, C.I = 95%
Standard deviation, = 0.90 hours
Sample size, n =?
Sample size =
But at 95% level of confidence, z = 1.96 from tables
Hence,
n = = 311.1696
Therefore, the required sample size is 312.
Q3
We are given:
X=120
N=200
Sample proportion, p= = =0.6
Confidence interval of a population proportion = z
95% confidence interval = 0.61.96*
95% confidence interval =0.60.0679
95% confidence interval = (0.5321, 0.6679)
Yes. The proportion of the Georgetown country residents who believe that country’s real estate taxes are too high is between 0.5321 and 0.6679 at 95% level of confidence.
Q4
First, state the null and alternative hypothesis
: (No significant difference in the mean selection sales)
: (Atleast one of the means of the selection sales is different)
Where;
= the mean for soft drink selection sales
= the mean for new registers sales
=the mean for dairy selection sales
To test where a significant difference exists, a one-way ANOVA model is conducted using Ms-Excel.
Anova: Single Factor |
|
|
|
|
|
|
|
|
|
|
|
|
|
SUMMARY |
|
|
|
|
|
|
Groups | Count | Sum | Average | Variance |
|
|
Column 1 | 40 | 11.5 |
|
| ||
Column 2 | 35 | 6.5 |
|
| ||
Column 3 | 43 | 8.6 | 4.3 |
|
| |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
ANOVA |
|
|
|
|
|
|
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 6.5333333 | 3.2666667 | 0.439462 | 0.65435 | 3.885294 | |
Within Groups | 89.2 | 12 | 7.4333333 |
|
|
|
|
|
|
|
|
|
|
Total | 95.733333 | 14 |
|
|
|
|
Decision rule at 5% significance level: Reject the null hypothesis if F> F critical (Kreyszig, 2010).
F=0.439
F critical = 3.885
Since F observed is less than F critical, we fail to reject the null hypothesis since we do not have enough evidence. Therefore we conclude that there is no significant difference in the mean selection of Coca-Cola stacked at four locations of the store.
Reference
Kreyszig, E. (2010). Advanced Engineering Mathematics, 10th Edition. John Wiley & Sons.