Standard deviation = 1.5 hours
Mean = 6.5 hours
Sample size = 100
Now the interval estimate is given as:
Α | 1-α | z |
.10 | .90 | 1.645 |
.05 | .95 | 1.96 |
.01 | .99 | 2.575 |
So the interval estimate is given as:
= 1.5
= 0.0866
Now the interval estimate is given as:
Standard deviation = 1.25 hours
Mean = 4 hours
Normal score,
For z ≤ 0.8, probability = 78.81%.
For z>0.8, probability = 21.19%.
For z > 1.6, probability = 5.48%.
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
Revenue = 27.7179 – 0.6943*Temperature
Putting Temperature = 38̊ C.
We get, Revenue = 1.3345 (in $00’s).
Hence, Revenue = $133.45.
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.
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).
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.