Statistics assignment solution
Solution for question 1
i) Steps for the drawing the histogram is given as follows:
First we have to make 7 classes for producing this histogram. Therefore first we find the minimum and range for the given data for the year 2005. Then we divide the range by the number 7 and then we add this value in a minimum value. This is the first value for our bin range. Now for the next value do the same.
After doing this, click on Data à then click on Data analysis à then click on histogram à then fill the range for data and bin range à press or enter ‘OK’.
The frequency chart for the year 2005 is given as follow:
|
Range |
Frequency |
|
6.56 |
3 |
|
7.51 |
11 |
|
8.46 |
19 |
|
9.41 |
22 |
|
10.36 |
21 |
|
11.31 |
1 |
|
12.26 |
3 |
|
More |
0 |
The histogram for the year 2005 is given as follow:
ii) Do the same steps for finding the frequency table and histogram for the year 2010.
The frequency table for the year 2010 is given as below:
|
Range |
Frequency |
|
4.24 |
1 |
|
5.27 |
4 |
|
6.31 |
12 |
|
7.34 |
19 |
|
8.38 |
31 |
|
9.41 |
8 |
|
10.44 |
4 |
|
More |
1 |
The histogram for year 2010 is given as below:
iii) The descriptive statistics for the year 2005 is given as below:
|
Year 2005 |
|
|
|
|
|
Mean |
8.738116938 |
|
Standard Error |
0.140952601 |
|
Median |
8.673027792 |
|
Mode |
#N/A |
|
Standard Deviation |
1.260718393 |
|
Sample Variance |
1.589410865 |
|
Kurtosis |
0.20566097 |
|
Skewness |
0.118221027 |
|
Range |
6.647425372 |
|
Minimum |
5.61111514 |
|
Maximum |
12.25854051 |
|
Sum |
699.049355 |
|
Count |
80 |
|
Largest(1) |
12.25854051 |
|
Smallest(1) |
5.61111514 |
|
Confidence Level(95.0%) |
0.28055913 |
iv) The descriptive statistics for the year 2010 is given as below:
|
Year 2010 |
|
|
|
|
|
Mean |
7.339501849 |
|
Standard Error |
0.146200934 |
|
Median |
7.430580714 |
|
Mode |
#N/A |
|
Standard Deviation |
1.307660909 |
|
Sample Variance |
1.709977053 |
|
Kurtosis |
0.612643421 |
|
Skewness |
-0.291526802 |
|
Range |
7.241837565 |
|
Minimum |
3.203132021 |
|
Maximum |
10.44496959 |
|
Sum |
587.1601479 |
|
Count |
80 |
|
Largest(1) |
10.44496959 |
|
Smallest(1) |
3.203132021 |
|
Confidence Level(95.0%) |
0.291005675 |
v) The comparison of performance and risk:
For the year 2005, the mean is higher than year 2010 and standard deviation for the year is lower than the year 2010. That is for the year 2005 there was a good performance and low risk than the year 2010.
Question 2 Solutions:
|
i)
The 99% confidence interval for Population average orders is given as below: |
|
|
Data |
|
| Sample Standard Deviation |
15.70044372 |
| Sample Mean |
52.77 |
| Sample Size |
73 |
| Confidence Level |
99% |
|
Intermediate Calculations |
|
| Standard Error of the Mean |
1.837597944 |
| Degrees of Freedom |
72 |
| t Value |
2.6459 |
| Interval Half Width |
4.8620 |
|
Confidence Interval |
|
| Interval Lower Limit |
47.91 |
| Interval Upper Limit |
57.63 |
ii) Solution:
Here we have to use the one sample t test for the population mean. Here we have to use the upper tailed test or right tailed test. The null and alternative hypothesis for this test is given as below:
Null hypothesis: H0: µ = 50 versus alternative hypothesis: Ha: µ > 50.
The calculations and decision for this test are given in the following table.
| One sample t test for population mean ( Right tailed – one tailed ) ( Upper tailed test) | |
|
Data |
|
| Null Hypothesis m= |
50 |
| Level of Significance |
0.01 |
| Sample Size |
73 |
| Sample Mean |
52.77 |
| Sample Standard Deviation |
15.70044372 |
|
Intermediate Calculations |
|
| Standard Error of the Mean |
1.8376 |
| Degrees of Freedom |
72 |
| t Test Statistic |
1.5086 |
|
Upper-Tail Test |
|
| Upper Critical Value |
2.3793 |
| p-Value |
0.0679 |
|
Do not reject the null hypothesis |
|
|
Calculations Area |
|
| For one-tailed tests: | |
| TDIST value |
0.067895 |
| 1-TDIST value |
0.932105 |
Question 3: Solution:
Here we have to use the two sample t test for comparing the two population means. The calculations for this test are given as below:
| Two sample t test | |
| (assumes unequal population variances) | |
|
Data |
|
| Hypothesized Difference |
0 |
| Level of Significance |
0.05 |
|
Population 1 Sample |
|
| Sample Size |
78 |
| Sample Mean |
0.766251524 |
| Sample Standard Deviation |
0.1332 |
|
Population 2 Sample |
|
| Sample Size |
65 |
| Sample Mean |
0.709795041 |
| Sample Standard Deviation |
0.1635 |
|
Intermediate Calculations |
|
| Numerator of Degrees of Freedom |
0.0000 |
| Denominator of Degrees of Freedom |
0.0000 |
| Total Degrees of Freedom |
123.0942 |
| Degrees of Freedom |
123 |
| Standard Error |
0.0253 |
| Difference in Sample Means |
0.0565 |
| Separate-Variance t Test Statistic |
2.2337 |
|
Lower-Tail Test |
|
| Lower Critical Value |
-1.6573 |
| p-Value |
0.9863 |
|
Do not reject the null hypothesis |
|
|
Calculations Area |
|
| Pop. 1 Sample Variance |
0.0177 |
| Pop. 2 Sample Variance |
0.0267 |
| Pop. 1 Sample Var./Sample Size |
0.0002 |
| Pop. 2 Sample Var./Sample Size |
0.0004 |
| For one-tailed tests: | |
| TDIST value |
0.0137 |
| 1-TDIST value |
0.9863 |
|
t-Test: Two-Sample Assuming Unequal Variances |
|
|
|
|
|
|
|
|
Reaction time on phone |
Reaction time not on phone |
|
Mean |
0.766251524 |
0.709795041 |
|
Variance |
0.017749911 |
0.026733018 |
|
Observations |
78 |
65 |
|
Hypothesized Mean Difference |
0 |
|
|
Df |
123 |
|
|
t Stat |
2.233663341 |
|
|
P(T<=t) one-tail |
0.01365628 |
|
|
t Critical one-tail |
1.657336398 |
|
|
P(T<=t) two-tail |
0.02731256 |
|
|
t Critical two-tail |
1.97943866 |
|
Question 4: Solution:
Here we cannot use the infer about the average value of the shareholdings are changed because for the using the one sample z test, it is important to know the population standard deviation and sample mean and hypothesis value. Or for using t test, it is important to know the sample mean and sample standard deviation, so we cannot use the t test here. That is, in brief, we do not infer about the average value of the shareholdings due insufficient information provided within problem.
