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
5
40
8
11.5
Column 2
5
35
7
6.5
Column 3
5
43
8.6
4.3
ANOVA
Source of Variation
SS
df
MS
F
P-value
F crit
Between Groups
6.5333333
2
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.
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