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This report looks at two datasets to answers different questions relating to them. The first dataset relates to a 2017 survey conducted across Australian campuses to get relevant information on key parameters that include: gender, birthday, GPA, pulse rate, the child rank in the family, the number of children in the family, the number of speed tickets, whether he/she is a smoker or not, the number of hours sleep per day, region, whether he/she lives on campus or not and whether he/she is over twenty years old. Using this data we present out findings on
The second dataset relates to a sample of 50 houses each in Suburb 1 (Newton) and Suburb 2 (Hurstville)in Australia. We investigate if the difference in average rents in these suburbs is statistically significant.
We first look at the data on GPA.
Mean | 3.243051 |
Standard Error | 0.029442 |
Median | 3.3 |
Mode | 3.4 |
Standard Deviation | 0.487344 |
Sample Variance | 0.237504 |
Kurtosis | 2.178944 |
Skewness | -1.0878 |
Range | 3 |
Minimum | 1 |
Maximum | 4 |
Sum | 888.596 |
Count | 274 |
From the data we get 274 sample values as some values are missing and need to be dropped. The mean GPA is 3.24, while the mode and median are both greater than mean. As the table above for descriptive statistics shows there is a small negative skewness in data. This is confirmed by the histogram drawn below. Most of the GPA values are concentrated y=towards the higher end of the range. The frequency of GPA within 3.25 and 3.5 is highest with 68 students scoring in this bracket. The minimum GPA is 1 but there are only 7 students who score less than 2.25 as GPA.
class interval | Frequency |
<1.25 | 1 |
1.25 - 1.5 | 2 |
1.5 - 1.75 | 0 |
1.75 - 2 | 3 |
2 - 2.25 | 1 |
2.25 - 2.5 | 17 |
2.5 -2.75 | 18 |
2.75 - 3 | 40 |
3 - 3.25 | 41 |
3.25 -3.5 | 68 |
3.5 - 3.75 | 46 |
>3.75 | 37 |
When we examine what dates the birthdays of students fall we see that the most birthdays fall on Wednesday. The proportion of such students with a birthday on Wednesday is 50/280 =0.178 or 17.8%. At the other extreme Friday sees least birthday- only 10% students having a birthday on Friday.
Moving to the rank of the students in the family, we have a maximum rank of 7, and rank 1 as the smallest rank. There is a dominance of students with rank 1- 88 out of 280 get rank 1. While only 1 students reported the rank 7. Rank 6 is equally scarce with just 2 students reporting it. As the rank rises. We have lesser students reporting it.
Lastly we come to the question of changes in rural population in the survey. As per our data 50 students report a rural background or 50/280 = 0.178. this is the sample proportion. The 2016 figure was 0.2. To test if this rural proportion has reduced in a statistically significant way we use a hypothesis test.
HO: null hypothesis: p=.2
H1; alternative hypothesis : p <.2
We use a z test here. The critical z value at 95% confidence level will be -1.645.
The test value = (sample proportion – hypothesized proportion/ SE )
The standard error = ( .2*.8/280)^.5 = 0.0239
The test value = ( 0.178 - 0.2)/.0239 = -0.00053
As the test value is less than critical value in absolute terms we DO NOT REJECT the null hypothesis. There is NO statistical evidence that the proportion of rural background students has decreased between 2016 and 2017.
We now use the technique of hypothesis testing to determine
we use Excel to look at difference in rents. Using the DATA ANALYSIS tab we conduct a z test for difference in mean. A z test is appropriate as the sample size exceeds 30 for both.
Let us consider the first question and formulate an appropriate hypothesis.
HO: null hypothesis: µ= 450
H1: alternative hypothesis : µ > 450
We use a z test here. The critical z value at 95% confidence level will be 1.645.
Sample average = 720
The test value = (sample average – hypothesized average/ SE )
The standard error = (standard deviation / 280 ^.5) = 124.3309/280^.5 = 7.439
The test value = ( 720-450)/7.439 =36.33
As the test value is more than critical value in absolute terms we DO NOT ACCEPT the null hypothesis. There is statistical evidence that the average rents in Newton exceed $450.
We now turn to the second question.
Ho: µ1 = µ2
H1: µ1 > µ2
| Suburb 1 (Newtown) |
Mean | 720 |
Known Variance | 15458.16 |
Observations | 50 |
Hypothesized Mean Difference | 0 |
z | 11.43148321 |
P(Z<=z) one-tail | 0 |
z Critical one-tail | 1.644853627 |
P(Z<=z) two-tail | 0 |
z Critical two-tail | 1.959963985 |
Using a 1 tail (right tail) test we can see that the z test value is 11.43, while the critical value at 95% level of confidence is 1.645. As test value > critical value we DO NOT ACCEPT the null hypothesis. There is statistical evidence that the rents in Newton are higher than in Hurstville.
Based on the datasets given we can make the following conclusions:
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