1. Most of the information technology developers claim that wireless connection would give speed of at least 11 Mbps compared to wired connection. A sample of 105 computers using wireless connection shows it gives a mean of 11.7 Mbps and a standard deviation of 2.3 Mbps. a) Do you think that the wireless usage should be implemented? Test at 7% significance level). b) Repeat the test by using p-value approach. 2. A jack is usually used for raising the axle of a motor so that a wheel may be changed. It is known that the weight of the vehicle that be raised by a jack is more than 100kg. a random sample of 125 jacks with a mean of 102.2kg and standard deviation of 15.17kg is taken. a) Do you agree that a jack can raise more than a 100 kg vehicle? (use α = 0.03) b) Repeat the test by using p-value approach. 3. In a normal situation, on average, a song needs a duration of at most 3.5 minutes to be played. The duration of a random sample of 11 songs is taken, and the data are as follows: 5.32 6.53 4.52 3.20 3.39 4.00 4.10 3.15 3.47 4.01 2.33 a) By using the data, do you agree with the statement that a song needs a duration of at most 3.5 minutes to be played? (use α = 0.05) b) Repeat the test by using p-value approach. 4. A researcher needs on average, 1.2 years to complete his short-term research. A random sample of 12 researcher are taken, and the duration (in years) to complete their research are recorded as below: 4.3 1.0 1.5 1.6 1.2 0.8 2.1 2.4 0.7 1.3 0.4

a) Do these data support the stamen that a researcher needs exactly 1.2 years to complete his research? (use α=0.05) b) Repeat the test by using p-value approach. 5. Newly purchased automobile tyres of a certain type are supposed to be filled with a pressure of at most 34 psi. A random sample of 36 such tyres was selected and it gives a sample mean of 34.66 psi and a standard deviation of 2.14 psi. a) Does this sample prove that such tyres fill...

...BUSINESSSTATISTICS
Agenda
Introduction
Descriptive Statistics
One–Tailed Test About a Population Mean
t-Test Using Two Independent Samples
t-Test For Paired Samples
Analysis of Variance (ANOVA)
Regression Analysis
U.S. Patent and Trademark Office (PT0)
Descriptive Statistics
Frequency Vs Total Provisional Filings
One–Tailed Test About a Population Mean
Selected a small sample of 12 months’ provisional application filings for large US
entities (for fiscal year 1996)
1. Determine the Hypotheses
H0: µ ≤ 600
Ha: µ > 600
2. Specify the level of significance
α = 0.05
3. Compute the value of the test statistic
4. Compute the p-value
5. Determine whether to reject H0
The P-value of the hypothesis test, equals 0.68. Since this is greater than the
specified significance level α = 0.05, the null hypothesis can not be rejected.
Conclusion : The data does not provide sufficient evidence to conclude that the
mean filings of provisional applications by large U.S. entities is greater than 600.
t-Test Using Two Independent Samples
Let µ1 denote the mean provisional applications filed by US large entities in 1996,
and µ2 denote the mean provisional applications filed by US large entities in 1997.
1. Determine the Hypotheses
H0: µ1 = µ2
Ha: µ1 < µ2
2. Specify the level of significance
α = 0.05
3. Compute the value of the test statistic
4. Compute the p-value
5. Determine whether to reject H0...

...3.009 kg. Using a 0.05 significance, conduct a hypothesis test to determine if the population mean weight of the cabin bags is greater than 3kg?
What is the null and alternative hypothesis for this test?
A H0: µ ≥ 3, H1: µ < 3
B H0: µ ≤ 3, H1: µ > 3
C H0: µ = 3, H1: µ ≠ 3
D H0: µ ≠ 3, H1: µ = 3
E None of the above
Ans: B
What is (are) the critical value(s)?
A -1.96 and 1.96
B -1.645
C -1.645 and 1.645
D 1.645
E None of the above
Ans: D
What is the calculated teststatistic?
A 2.12
B 1.43
C 1.78
D 1.23
E None of the above
Ans: A
What is your conclusion? (choose the most correct answer)
A Reject Ho at the 5% level of significance.
B Accept Ho at the 5% level of significance.
C Not sure.
D None of the above.
E Need more information.
Ans: A
Calculate the p-value
A P value = 0.017
B P value = 0.09
C P value = 0.05
D Not sure
E None of the above.
Ans: A
Interpret the meaning of the p-value
A The probability of observing a test statistic more extreme than the observed sample value given the null hypothesis is true.
B The observed level of significance.
C The smallest value of α for which H0 can be rejected.
D All of the above
E None of the above
Ans: D
Problem 2 (Multiple Regression):
A developer who specialises in holiday cottage properties is considering purchasing a large tract of land adjoining a lake. The current owner of the tract has already subdivided the land into...

...that we are accepting the alternative hypothesis and this statement works vice versa. In this case that means that the null hypothesis can be rejected or disproving. For the data set that was given the null hypothesis also known as H-nought was µ1=µ2, while the alternative hypothesis is µ1<µ2. Null hypothesis states that the amount of rural nurse homes was equal to the average amount of beds used. Alternative hypothesis states that rural area nursing homes uses fewer amounts of beds. The claim indicated to what kind of test was going to be used and since I claimed that the rural area were going to have a lower average number of beds it states that the shaded area on the critical value test will be less than zero.
Table 1. Descriptive statistics for the given null and alternative hypothesis that includes the sample, mean, median, standard deviation, maximum values, and minimum values.
Sample Size
Mean
Median
Standard Deviation
Maximum Value
Minimum Value
Rural Area
34
0.6538
1.0000
0.4803845
1
0
Bed
4850
93.27
88.00
40.85273
244.00
25.00
Figure 1. This figure illustrates the critical value test for the left-tailed test. The critical value that was needed for the test was -1.692 according to the t-table since our sample size was 34. Used the degree of freedom formula to find the critical value.
Figure 2. This figure reflects to the p-value. When we figured out the p-value we used pt(t,33). Since pt(t,33) equaled 0.0137855 that indicated...

...of 1000 flights and proportions of three routes in the sample. He divides them into different sub-groups such as satisfaction, refreshments and departure time and then selects proportionally to highlight specific subgroup within the population. The reasons why Mr Kwok used this sampling method are that the cost per observation in the survey may be reduced and it also enables to increase the accuracy at a given cost.
TABLE 1: Data Summaries of Three Routes
Route 1
Route 2
Route 3
Normal(88.532,5.07943)
Normal(97.1033,5.04488)
Normal(107.15,5.15367)
Summary Statistics
Mean
88.532
Std Dev
5.0794269
Std Err Mean
0.2271589
Upper 95% Mean
88.978306
Lower 95% Mean
88.085694
N
500
Sum
44266
Summary Statistics
Mean
97.103333
Std Dev
5.0448811
Std Err Mean
0.2912663
Upper 95% Mean
97.676525
Lower 95% Mean
96.530142
N
300
Sum
29131
Summary Statistics
Mean
107.15
Std Dev
5.1536687
Std Err Mean
0.3644194
Upper 95% Mean
107.86862
Lower 95% Mean
106.43138
N
200
Sum
21430
From the table above, the total number of passengers for route 1 is 44,266, route 2 is 29,131 and route 3 is 21,430 and the total numbers of passengers for 3 routes are 94,827.
Although route 1 has the highest number of passengers and flights but it has the lowest means of passengers among the 3 routes. From...

...BUS 105e:
Statistics
By Dr Tony Halim
GBA: 27 February 2013
Done by:
Koh En Song Andrew (Q1211397)
Melissa Teo Kah Leng (E1011088)
Woon Wei Jie Jared
T 04
1.
Over the span of 100 days, the total revenue for Unicafe North and Unicafe West is $21876.60 and $22042.00 respectively. The average revenue for Unicafe North is $218.77. The average revenue for Unicafe West is $220.42. The highest revenue occurred on the 88th day for both outlets. The lowest revenue occurred on 39th day for both outlets. Generally, both outlets earn roughly the same amount of revenue each day.
2a.
Confidence interval is a range of values constructed from sample data so that the population parameter is likely to occur within that range at a specific probability (Lind, Marchal & Wathen, 2013).
Using the 95% level of confidence, the confidence interval for Unicafe West is 220.42 6.211. The confidence interval limits are $214.21 and $226.63 (rounded off to 2 decimal places).
Using the 95% level of confidence, the confidence interval for Unicafe North is 218.766 5.571. The confidence interval limits are $213.20 and $224.34 (rounded off to 2 decimal places).
In the event that Mr Yeung wants to predict his potential revenue for the next one hundred days, 95% of the confidence intervals would be expected to contain the population mean. The remaining 5% of the confidence intervals would not contain the population mean, average revenue earned per day....

...3ER PARCIAL
Inferential statistics
Sampling
* The purpose of sampling is to select a set of elements (sample) from a population that we can use to estimate parameters about the population
* The bigger the sampling, the more accurate our parameters will be.
example:
In the experiment of deciding if CEGL girls are smarter that CEGL boys, which would be your statistical hypothesis?
Hypothesis testing
But now, you already gathered information about a sample
No, you will test if your hypothesis are true or not
Hypothesis testing involves testing the difference between a hypothesized value of a population parameter and the estimate of that parameter, calculated from the sample
example:
If you want to know if CEGL girls are smarter that CEGL boys, you ask a few girls/boys their grades and compare averages, we will use Excel to compare the population and sample means. If the difference is too high, we can’t compare.
In statistics, the hypothesis to be tested is called “null hypothesis” and has the symbol “Ho”
The other option of the hypothesis is the “alternative hypothesis” and its symbol is “Ha”
1 Ho: “There is no difference between (independent variable) and (dependent variable)”
2 Ha: “There is a difference between (independent variable) and (dependent variable)”
example:
In the experiment of deciding if CEGL girls are smarter that CEGL boys, which would be your statistical hypothesis?
Ho: There is no...

...Case Problem 1: National Health Care Association(Descriptive Statistics)
The National Health Care Association is concerned about the shortage of nurses the health care profession is projecting for the future. To learn the current degree of job satisfaction among nurses, the association has sponsored a study of hospital nurses throughout the country. As part of this study, a sample of 50 nurses was asked to indicate their degree of satisfaction in their work, their pay and their opportunities for promotion. Each of the three aspects of satisfaction was measured on a scale from 0 to 100, with larger values indicating higher degrees of satisfaction. The data collected also showed the type of hospital employing the nurses. The types of hospitals were private (P), Veterans Administration (VA) and University (U). The complete data set is on the file named “Health”.
Managerial Report:
Use methods of descriptive statistics to summarize the data. Present the summaries that will be beneficial in communicating the results to others. Discuss your findings. Specifically, comment on the following questions.
1. On the basis of the entire data set and the three job satisfaction variables, what aspect of the job is most satisfying for the nurses? What appears to be the least satisfying? In what area(s), if any, do you feel improvements should be made? Discuss
2. On the basis of descriptive measures of variability, what measure of job...

...significantly from the 24 hours?) (a) Use the steps of hypothesis testing. (b) Sketch the distributions involved. (c) Explain your answer to someone who has never taken a course in statistics.
Solution:
(a) Use the steps of hypothesis testing.
Size of sample, n = 8
Degree of freedom = n-1 = 8-1 = 7
Sum of sample = i=1∑n=8xi = (25+27+25+23+24+25+26+25) = 200
| Time (in Hours) | Sum | Mean(xm) | (xi- xm)2 | Standard Deviation |
1 | 25 | 200 | Mean = Sum/n= 200/8= 25 | 0 | σ = √( i=1∑n=8(xi-xm)2/(n-1))= √(10/7)= √1.4285= 1.195 |
2 | 27 | | | 4 | |
3 | 25 | | | 0 | |
4 | 23 | | | 4 | |
5 | 24 | | | 1 | |
6 | 25 | | | 0 | |
7 | 26 | | | 1 | |
8 | 25 | | | 0 | |
Here X = 25
σ = 1.195
We can define two-tailed statistics for above observations as follows:
Null Hypothesis: H0: =24 vs. Ha: 24
Rejection region:
z < -z/2 or z > z/2
Here significance level is 0.05,
So, z0.025 = 1.96 (Using Statistical Ratio Calculator
Now,
z = (X - μ) / σx
Where X is a normal random variable, μ is the mean, and σ is the standard deviation.
Where n is the sample size.
Calculating t-test statistics:
z = 25-24/(1.195/√8) = 1* √8/1.195= 2.828/1.195 = 2.366
Calculating p-value:
Degree of freedom = DF = 8-1 = 7
Absolute value of calculated t-test statistics = |z| = |2.366| = 2.366
P(|z7| < 2.366) = 0.049899
The p-value of 4.9% is very slightly less than...