ECO502 Sample Assignment
- 1(a) Given:
Standard deviation = 1.5 hours
Mean = 6.5 hours
Sample size = 100
- According to the 99 % population mean, standard error mean is given as :
Now the interval estimate is given as:
- From the Table of reliability coefficient:
Α | 1-α | z |
.10 | .90 | 1.645 |
.05 | .95 | 1.96 |
.01 | .99 | 2.575 |
So the interval estimate is given as:
- 5
- 8862, 6.1137
- According to 95 % population means, SEM is given as:
= 1.5
= 0.0866
Now the interval estimate is given as:
- Mean
- 5
- 6697, 6.3302
- Interval estimates are affected by the sample size and confidence interval. When the confidence interval is decreased and the sample size is increased then the upper limit of the interval estimate is decreased while the lower limit of the interval estimate is increased. A higher confidence interval shows a range which is closer to the mean of the sample, while a lower confidence interval covers up a more dispersed range of the sample.
- 1 (b) Given:
Standard deviation = 1.25 hours
Mean = 4 hours
- X = 5 hours
Normal score,
For z ≤ 0.8, probability = 78.81%.
For z>0.8, probability = 21.19%.
- For z ≤ 1.6, probability = 94.52%.
For z > 1.6, probability = 5.48%.
- Probability of one person playing more than 5 hours = 21.19% = 0.2119.
Probability of four persons all playing for more than 5 hours = (0.2119)^{4}.
2a)
N = 25
H_{0}: The lecture recordings did not produce any changes.
H_{1}: The lecture recordings changed the scores of students.
When the lecture recordings were provided,
Since,
Hence, we can reject the null hypothesis.
2b)
N = 50
H_{0}: The number of additional hours of sleep is less than 2 hours.
H_{1}: The number of additional hours of sleep is at least 2 hours.
Now, when the pills are provided,
Since, , hence we may not reject the null hypothesis.
3a)
2003, A = 25%, B – 20%, C-10%, D – 45%.
Taking sample as 1200, A=300, B=240, C=120, D=540.
2004, A=280, B=270, C=90, D=560.
H_{0}: The market has not changed since 2003.
H_{1}: The market has changed since 2003.
And,
Now since, , so the null hypothesis can be rejected inferring that the market has changed since 2003.
3b)
Insurance Preference | |||
Gender | Term | Whole life | No insurance |
Female | 170 | 110 | 470 |
Male | 195 | 75 | 480 |
H_{0} = Preference is same.
H_{1} = Preference is different.
Expected,
Term | Whole life | No insurance | ||||
Gender | Observed | Expected | Observed | Expected | Observed | Expected |
Female | 170 | 182.50 | 110 | 92.50 | 470 | 475 |
Male | 195 | 182.50 | 75 | 92.50 | 480 | 475 |
Significance level = 5% = 0.05
H_{0} will be rejected if
Here,
= 8.4392
Now since, , hence the null hypothesis may be rejected.
3c)
Null hypothesis, H_{0} = The given data follows a normal distribution.
Alternate Hypothesis, H_{1} = Given data is not normally distributed.
Mean = 26.80; Standard deviation = 6.378
Significance level = 5% = 0.05
H_{0} will be rejected if
Dividing the sample into 4 intervals based on deviations:
Standardized intervals | Selected intervals | Probability | Expected | Observed | Chi-score |
Z ≤ -1 | X ≤ 20.422 | 0.1587 | 6.348 | 4 | 0.8685 |
-1 < Z ≤ 0 | (20.422, 26.80] | 0.3413 | 13.652 | 20 | 2.9517 |
0 < Z ≤ 1 | (26.80, 33.178] | 0.3413 | 13.652 | 11 | 0.5152 |
1 < Z | X > 33.178 | 0.1587 | 6.348 | 5 | 0.2862 |
Now, from the table,
Now since, .
Thus, we can reject the null hypothesis. And may conclude the data to be not normally distributed
- a) From the graph, the regression may be obtained as:
Revenue = 27.7179 – 0.6943*Temperature
Putting Temperature = 38̊ C.
We get, Revenue = 1.3345 (in $00’s).
Hence, Revenue = $133.45.
- b) As it may be seen from the calculations, the slope of the equation between the revenue and temperature is negative and has interception at 38̊ C. Hence, the revenue obtained from the equation is obtained to be 1.3345.
But, seeing the data it may be observed that the sales for the day when temperature reached 35̊ C and 40̊ C were $450 and $350 respectively. Hence, it may be inferred that the above prediction is not reasonable.
- c)
Since, , so it may be inferred that there is a linear relationship between the price and odometer reading but the slope does not fits to the data pretty well because of the high value of the standard error observed (5.2027).
- d) The analysis carried over the given data yields the results as having an R-squared value of 0.7472 which should be even closer to 1 to yield the best results.
The standard error observed is 5.2027 and the mean observed for y values is 9.667.
The standard error is thus 53.82% of the mean value of the y axis. It therefore does not fits to the data well.
This model is thus providing a bad estimation of the prices and the sales data of the vendors. It is thus recommended to increase the sample size where the data may be collected twice a day.
Question 5:
Null Hypothesis, H_{0}: Students in different semesters do not produce different grades.
Alternate Hypothesis, H_{1}: Students in different semesters produce different grades.
Test statistics:
Significance level:
Decision rule:
Reject H_{0} if intervals differ in their estimations.
Value of the test statistics:
For the first class,
For the second class,
Conclusion: As it may be observed from the intervals obtained that the intervals for the two classes differ significantly and are distributed similarly over the interval ranges. Hence, we may reject the null hypothesis and may infer that the grades obtained by students in the two different semesters vary with respect to each other.
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