logo

Confidence Interval for Proportion and Mean of Kuku on Marlborough Sounds Mussel Farm

   

Added on  2022-10-17

7 Pages804 Words92 Views
161.120 Ass2 2019Name: _______A. Student______________ ID number: ______12345678__________________
161.120 Ass2 2019
Student’s name:
Student’s ID:
Page 1

161.120 Ass2 2019Name: _______A. Student______________ ID number: ______12345678__________________
Part A: Confidence Interval for a proportion [15 marks]
1. Proportion
The proportion of kuku that are female in the sample is 0.5
2. Confidence interval
General formula for a CI: CI = p̂ ± z * (p̂(1 - p̂)/n)
Sample size = 100
Statistic = 50
Standard Error = (p̂(1 - p̂)/n) = √(0.5*(1-0.5)/100) = 0.05
Critical value = 1.6449
Interval half width = 1.6449 * 0.05 = 0.0822
95% CI = (0.5–0.0822, 0.5+0.822) = (0.4178, 0.5822)
Interpretation: We are 95% confident that the proportion of female kuku in the
Marlborough Sounds mussel farm is between 41.78% and 58.22%.
3. Validity of the confidence interval
The confidence interval is valid.
This conclusion is since there is a confidence level (95% confidence level), there is a
statistic (50 or 0.5), and that there is a margin of error (0.0822).
4. Conclusion
Hypothesis
H0: Proportion of male kuku is more than 40%
H1: Proportion of male kuku is less than or equal to 40%
General formula for a CI: z = (P - p̂)/SE
Sample size = 100
Statistic = 50
Standard Error = (p̂ (1 - p̂)/n) = √ (0.5*(1-0.5)/100) = 0.05
Critical value = 1.6449
Page 2

161.120 Ass2 2019Name: _______A. Student______________ ID number: ______12345678__________________
Z sample statistic = (0.4-0.5)/0.05
p-value = 0.02
Decision: Reject the null hypothesis
Conclusion: The survival of Marlborough Sounds mussel farm is not in threat since the
population of the male kuku is less than 40%.
Part B: Confidence interval for a mean
1. Exploratory data analysis
a)
b)
The distribution is normally distributed since it is bell shaped. Conversely, the
distribution is positively skewed since its tail is on the left side.
c)
Mean = 98.390
Median = 93.500
Comment: It is evident that the distribution is positively skewed since the mean is
grater than the median.
Page 3

End of preview

Want to access all the pages? Upload your documents or become a member.

Related Documents
Introductory Biostatistics Assignment 2: Inference
|5
|1422
|373

Strategics for business
|13
|2513
|61

Statistics Assignment - Substantive Significance
|8
|740
|68

Statistics Study Material
|7
|1378
|58

Statistical Methods For Rates and Proportions
|6
|456
|16

ECON1095 Quantitative Methods In Finance
|13
|925
|171