Statistics questions need in 1 hour
115 FINAL EXAM PART II
SHOW ALL WORK FOR FULL CREDIT
1. What is the area to the right Z = -1 in the normal distribution?
2. Given X ~ N(69.5, 2.4), find P( 62 < x < 73).
3. As Mario was writing, he measured his pulse rate to be 48 beats per minute. Assume that a large sample of adult males has pulse rates with a mean of 67.3 beats per minute and a standard deviation of 10.3 beats per minute. Is Mario’s pulse rate unusual?
4. A Pew Research Center poll of 1007 randomly selected U.S. adults showed that 85% of the respondents know what Twitter is. Find the 95% confidence interval estimate of the population proportion p.
5. Twelve different video games showing substance use were observed. The duration times (in seconds) of alcohol use were recorded, with the times listed below. The design of the study justifies the assumption that the sample can be treated as a simple random sample.
a) Use the sample data to construct a 90% confidence interval estimate of µ, the mean duration time that the video showed the use of alcohol.
b) Interpret the confidence interval in context of the problem.
84, 14, 583, 50, 0, 57, 207, 43, 178, 0, 2, 57
6. A researcher is testing the claim that the average adult consumes 1.7 cups of coffee per day. Assume the population standard deviation is 0.5 cups per day. A sample of 25 adults averaged 1.85 cups of coffee per day. Test the researcher’s claim at the 5% significance level.
7. A researcher claims that the average cost for a family of four to attend a Major League Baseball game is not equal to $172. A random sample of 22 families reported an average cost of $
18
9.34, with a sample standard deviation of $33.65. Assume the population is normally distributed. Test the claim at the 10% significance level.
8. The table below refers to the average bill per customer at a restaurant when different types of background music were played. The managers would like to determine the impact music has on the size of the bill. Assume the population is normally distributed.
Fast Music |
Slow Music |
|
Sample Mean |
$39.65 |
$42.60 |
Sample Size |
18 |
23 |
Population standard deviation |
$4.21 |
$5.67 |
Test the hypothesis that the average bill of customers exposed to fast music is less than the average bill of customers exposed to slow music at the 5% significance level.
9. A study was conducted to test the effectiveness of a software patch in reducing system failures over a six-month period. Results for randomly selected installations are shown in the table below. The before and after values are normally distributed. Test at the 1% significance level. Interpret your conclusion in context of the problem.
10. A bank has recently acquired a new branch and thus has customers in this new region. They are interested in the default rate in their new region. They wish to test the hypothesis that the default rate is different from their current customer base. They sample 200 files in area A, their current customers, and find that 20 have defaulted. In area B, the new customers, another sample of 200 files shows 12 have defaulted on their loans. At the 10% significance level can we say that the default rates are different?
115 FINAL EXAM PART II
SHOW ALL WORK FOR FULL CREDIT
1. What is the area to the right Z = -1 in the normal distribution?
The area to the right of z=-1 is
1-0.1587 =0.8413
2. Given X ~ N(69.5, 2.4), find P( 62 < x < 73).
3. As Mario was writing, he measured his pulse rate to be 48 beats per minute. Assume that a large sample of adult males has pulse rates with a mean of 67.3 beats per minute and a standard deviation of 10.3 beats per minute. Is Mario’s pulse rate unusual?
The p value (0.030) is less than 0.05 so yes the Mario’s pulse rate unusual
4. A Pew Research Center poll of 1007 randomly selected U.S. adults showed that 85% of the respondents know what Twitter is. Find the 95% confidence interval estimate of the population proportion p.
N=1007
P=0.85
Q=1-0.85=0.15
Z0.05/2=1.96
95% confidence interval = P±ME
95% confidence interval=0.85±0.0221=(0.8280, 0.8721)
We are 95% confidence interval of the true proportion is between 0.8280 and 0.8721
5. Twelve different video games showing substance use were observed. The duration times (in seconds) of alcohol use were recorded, with the times listed below. The design of the study justifies the assumption that the sample can be treated as a simple random sample.
a) Use the sample data to construct a 90% confidence interval estimate of µ, the mean duration time that the video showed the use of alcohol.
b) Interpret the confidence interval in context of the problem.
84, 14, 583, 50, 0, 57, 207, 43, 178, 0, 2, 57
6. A researcher is testing the claim that the average adult consumes 1.7 cups of coffee per day. Assume the population standard deviation is 0.5 cups per day. A sample of 25 adults averaged 1.85 cups of coffee per day. Test the researcher’s claim at the 5% significance level.
H0 : μ ≤1.7
HA : μ >1.7
Mean=1.85
N=25
SD=0.5
Df=n-1=25-1=24
The p value corresponding to the tothe t value 1.5 and df=24 is 0.14667
Fail to Reject the H0 and no evidence to support the claim and conclude that the average adult consumes 1.7 cups or less of coffee per day
7. A researcher claims that the average cost for a family of four to attend a Major League Baseball game is not equal to $172. A random sample of 22 families reported an average cost of $
18
9.34, with a sample standard deviation of $33.65. Assume the population is normally distributed. Test the claim at the 10% significance level.
H0 : μ =172
HA : μ ≠172
Mean=189.34
N=22
SD=33.65
Df=n-1=22-1=21
The p value corresponding to the to the t value 2.4264 and df=21 is 0.0244
Reject the H0 and found evidence to support the claim and conclude that the average cost for a family of four to attend a Major League Baseball game is not equal to $172
8. The table below refers to the average bill per customer at a restaurant when different types of background music were played. The managers would like to determine the impact music has on the size of the bill. Assume the population is normally distributed.
Fast Music |
Slow Music |
|
Sample Mean |
$39.65 |
$42.60 |
Sample Size |
18 |
23 |
Population standard deviation |
$4.21 |
$5.67 |
Test the hypothesis that the average bill of customers exposed to fast music is less than the average bill of customers exposed to slow music at the 5% significance level.
Hypothesis:
Average bill of customers exposed to Fast Music= μ1
Average bill of customers exposed to Slow Music= μ2
H0 : μ1 =μ2
HA : μ1 <μ2
Df1=17
Df2=22
One tail test with the degree of freedom 39
The p value is 0.0363
Reject the H0 and found evidence to support the claim and conclude that the average bill of customers exposed to fast music is less than the average bill of customers exposed to slow music at the 5% significance level.
9. A study was conducted to test the effectiveness of a software patch in reducing system failures over a six-month period. Results for randomly selected installations are shown in the table below. The before and after values are normally distributed. Test at the 1% significance level. Interpret your conclusion in context of the problem.
10. A bank has recently acquired a new branch and thus has customers in this new region. They are interested in the default rate in their new region. They wish to test the hypothesis that the default rate is different from their current customer base. They sample 200 files in area A, their current customers, and find that 20 have defaulted. In area B, the new customers, another sample of 200 files shows 12 have defaulted on their loans. At the 10% significance level can we say that the default rates are different?
Solution
Here Pcurrent and Pnew are the two population proportions we want to test
P’current – P’new = the difference in the proportions of customers who defaulted
H0: Pcurrent = Pnew
Ha: Pcurrent ≠ Pnew
Pooled sample proportion = (20+12)/(200+200)
=0.08
Estimated proportion P’current=20/200=0.1
Estimated proportion P’new=12/200=0.06
Estimated difference =0.1-0.06
=0.04
The standard error is
=squareroot [0.08(1-0.08)(1/200+1/200)]
=0.0271293
Test statistics = (0.1-0.06)/0.0271293
=1.4744
At the 10% significance level, critical value of z is 1.645
Since the test statistic is less than the critical value we fail to reject the null hypothesis i.e. there is not enough evidence to support the claim that two default rates are different.