A psychology professor asked her sophomore students Does eit
A psychology professor asked her sophomore students \"Does either of your parents allow you to drink alcohol around him or her?\" and when drinking (not necessarily around your parent) \"How many drinks do you typically have per session?\" (For these purposes, a drink is defined as a 12 oz. beer, a 4oz. glass of wine, or a one oz. shot of liquor.)
The next tables contains the responses from female students who are not abstainers.
Drinks per session for students whose parent allows them to drink
Drinks per session for students whose parents do not allow them to drink
If you treat these students as an SRS, construct a 90% confidence interval for the population proportion of sophomore college students at least one of whose parents allows them to drink around him or her. Do both the large sample and the plus four procedure.
Large Sample:
Lower confidence level:
Upper confidence level:
Plus four method:
Lower confidence level:
Upper confidence level:
| 2.5 | 1 | 2.5 | 3 | 1 | 3 | 3 | 3 | 2.5 | 2.5 | 3.5 | 5 | 2 |
| 7 | 7 | 6.5 | 4 | 8 | 6 | 6 | 3 | 6 | 3 | 4 | 7 | 5 |
| 3.5 | 2 | 1 | 5 | 3 | 3 | 6 | 4 | 2 | 7 | 5 | 8 | 1 |
| 6 | 5 | 2.5 | 3 | 4.5 | 9 | 5 | 4 | 4 | 3 | 4 | 6 | 4 |
| 5 | 1 | 5 | 3 | 10 | 7 | 4 | 4 | 4 | 4 | 2 | 2.5 | 2.5 |
Solution
Answer to the question)
There are 60 values in the first set for students whose parents allow them to drink
the mean value of this data is 4.2167
the mean value of this data is 2.0654
.
Thus we got :
n = 60
x bar = 4.2167
s = 2.0654
.
The formula of Confidence interval is:
x bar - t* s/sqrt(n) , xbar + t* s / sqrt(n)
.
the value of t for 90% confidence interval and df = 59 is: 1.6711
.
On plugging the values we get
4.2167 - 1.6711 * 2.0654 /sqrt(60) , 4.2167 + 1.6711 * 2.0654 /sqrt(60)
3.7711 , 4.6623

