2a) a) Increasing the difference between the sample mean and the original. The z score represents the distance of each X or score from the mean. If the distance between the sample mean and the population mean the z score will increase.

b) Increasing the population standard deviation.
The standard deviation is the factor that is used to divide by in the equation. the bigger the SD, then the smaller the z score.
c) Increasing the number of scores in the sample.
Should bring the samples mean closer to the population mean so z score will get smaller.

4a) The boundaries for critical region, when alpha level is is changed from α = .05 to .01, the size of the critical region will increase in size and the boundaries will be closer to the center of the distribution. This will allow me to see if there is error or not and also accept or reject the null hypothesis. If the alpha level is changed from .05 to .01 a) what happens to the boundaries for the critical region?

It reduces the power of the test to prove the hypothesis.
You increase the chance of rejecting a true H
b) what happens to the probability of a type 1 error?
Type 1 error is falsely reporting a hypothesis,
Where you increase the chance that you will reject a true null hypothesis.

4b) The probability of a Type I error is determined by the alpha level. Since the alpha level has decreased, we can expect increase in error. 6a) The independent variable is the application of study –skills training program and the dependent variable is the standardized achievement test. 6b) 6) A researcher is investigating the effectiveness of a new study skills training program for elementary school childreen. A sample of n=25 third grade children is selected to participate in the program and each child is given a standardizrd achievement test at the end of year. For the regular population of third grade children, scores on the test form a normal distribution with a mean u = 150, and a standard deviation q = 25....

...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 StatisticsMean
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 StatisticsMean
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 StatisticsMean
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 the sample of 1000 flights,...

...appropriate descriptive statistics to summarize the training time data for each method. What similarities or differences do you observe from the sample data?
Descriptive analysis in excel has been used to come up with relevant figures of the given data samples which is tabulated below:
Descriptive Statistics | Current | Proposed |
Mean | 75.06557 | 75.42623 |
Standard Error | 0.505094 | 0.32091 |
Median | 76 | 76 |
Mode | 76 | 76 |
Standard Deviation | 3.944907 | 2.506385 |
Sample Variance | 15.5623 | 6.281967 |
Kurtosis | -0.06933 | 0.58694 |
Skewness | -0.22053 | -0.28749 |
Range | 19 | 13 |
Minimum | 65 | 69 |
Maximum | 84 | 82 |
Sum | 4579 | 4601 |
Count | 61 | 61 |
Analysis of descriptive statistics shows that both the current and the proposed plan have almost similar mean completion hours which stand at 75.06 and 75.42 for the current and proposed respectively. Both the plans have exact same median and mode. However, the standard deviation in the current plan (3.94) is higher than that in the proposed plan (2.56), which is ultimately leading to the higher variance in the current plan. This suggests that the completion hours are more dispersed the mean value in the current plan, hence the mean does not give the true picture of data distribution whereas in the proposed plan, data for completion hours is comparatively more congregated....

...Trajico, Maria Liticia D.
BSEd III-A2
REFLECTION
The first thing that puffs in my mind when I heard the word STATISTIC is that it was a very hard subject because it is another branch of mathematics that will make my head or brain bleed of thinking of how I will handle it. I have learned that statistic is a branch of mathematics concerned with the study of information that is expressed in numbers, for example information about the number of times something happens. As I examined on what the statement says, the phrase “number of times something happens” really caught my attention because my subconscious says “here we go again the non-stop solving, analyzing of problems” and I was right. This course of basic statistic has provided me with the analytical skills to crunch numerical data and to make inference from it. At first I thought that I will be alright all along with this subject but it seems that just some part of it maybe it is because I don’t pay much of my attention to it but I have learned many things. I have learned my lesson.
During our every session in this subject before having our midterm examination I really had hard and bad times in coping up with this subject. When we have our very first quiz I thought that I would fail it but it did not happen but after that, my next quizzes I have taken I failed. I was always feeling down when in every quiz I failed because even though I don’t like this...

...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...

...
MBA 501A – [STATISTICS]
ASSIGNMENT 4
INSTRUCTIONS: You are to work independently on this assignment. The total number of points possible is 50. Please note that point allocation varies per question. Use the Help feature in MINITAB 16 to read descriptions for the data sets so that you can make meaningful comments.
[10 pts] 1. Use the data set OPENHOUSE.MTW in the Student14 folder. Perform the Chi
Square test for independence to determine whether style of home and location are are related. Use α = 0.05. Explain your results.
Pearson Chi-Square = 37.159, DF = 3, P-Value = 0.000
Likelihood Ratio Chi-Square = 40.039, DF = 3, P-Value = 0.000
The P value associated with out chi square is 0.00 and the Alpha level is 0.05 so we reject the null hypothesis. The P- value is less than the alpha level. So, we conclude that style of homes and locations are not related.
[10 pts] 2. Use the data set TEMCO.MTW in the Student14 folder. Perform the Chi
Square test for independence to determine whether department and gender are related. Use α = 0.05. Explain your results.
Pearson Chi-Square = 1.005, DF = 3, P-Value = 0.800
Likelihood Ratio Chi-Square = 1.012, DF = 3, P-Value = 0.798
The P-value associated with out chi square is 0.800 and the Alpha level is 0.05 we can see that we are unable to reject the null hypothesis. The P- value is greater than the alpha level. So, we conclude that departments and gender are related..
[30 pts] 3. Use the data set...

...interval for the mean selling price of the home.
b) Develop a 95 percent confidence interval for the mean distance of the home is from the center of the city.
Exercise 02
a) A recent article in Bangladesh Observer indicated that the mean selling price of the homes in the city of Dhaka is more than Tk. 2200. Can we conclude that the mean selling price in the Gulshan area is more than Tk. 2200? Use the 0.01 significance level. What is the P-value?
b) The same article reported the mean size was more than 2100 square feet? Can we conclude that the mean size of homes sold in the Gulshan area is more than 2100 square feet? Use the 0.05 significance level. What is the P-value?
Exercise 03
a) At the 0.05 significance level, can we conclude that there is a difference in the mean selling price of homes with a pool and homes without a pool?
b) At the 0.05 significance level, can we conclude that there is a difference in the mean selling price of homes with an attached garage and homes without a garage?
Exercise 04
a) At the 0.01 significance level, is there a difference in the variability of the selling prices of the homes that have a pool versus those that do not have a pool?
b) At the 0.02 significance level, is there a difference in the variability of the selling prices of the homes with an attached garage versus...

...1.3.5.16.
Kolmogorov-Smirnov Goodness-of-Fit Test
Purpose:
Test for Distributional AdequacyThe Kolmogorov-Smirnov test (Chakravart, Laha, and Roy, 1967) is used to decide if a sample comes from a population with a specific distribution.
The Kolmogorov-Smirnov (K-S) test is based on the empirical distribution function (ECDF). Given N ordereddata points Y1, Y2, ..., YN, the ECDF is defined as
\[ E_{N} = n(i)/N \]
where n(i) is the number of points less than Yi and the Yiare ordered from smallest to largest value. This is a step function that increases by 1/N at the value of each ordered data point.
The graph below is a plot of the empirical distribution function with a normal cumulative distribution function for 100 normal random numbers. The K-S test is based on the maximum distance between these two curves.
Characteristics and Limitations of the K-S TestAn attractive feature of this test is that the distribution of the K-S test statistic itself does not depend on the underlying cumulative distribution function being tested. Another advantage is that it is an exact test (the chi-square goodness-of-fit test depends on an adequate sample size for the approximations to be valid). Despite these advantages, the K-S test has several important limitations:
1. It only applies to continuous distributions.
2. It tends to be more sensitive near the center of the distribution than at the tails.
3. Perhaps the most serious limitation is that the...

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