Confidence Intervals and Hypothesis TestingName:
Math 256, Nelson

1. A soft-drink machine is regulated so that the amount of drink dispensed is approximately normally distributed with a standard deviation equal to 1.5 deciliters. Find a 95% confidence interval for the mean of all drinks dispensed by this machine if a random sample of 36 drinks had an average content of 2.25 dl.

2. The heights of a random sample of 50 college students showed a mean of 174.5 centimeters. Assume the standard deviation is 6.9 cm. What can we assert with 98% confidence about the possible size of our error if we estimate the mean height of all college students to be 174.5?

3. How large a sample is needed in #1 above is we wish to be 95% confident that our sample mean will be within 0.3 dl of the true mean?

4. A random sample of 1000 recorded deaths in the United States during the past year showed an average life span of 71.8 years. Assume a standard deviation of 8.9 years. Find a 95% confidence interval for the actual average life span.

5. On a certain national school test [pic]and [pic]. A principal claims that her students are above average. She takes a random sample of 75 and finds a mean of 525. Is her claim justified?

6. A random sample of 100 recorded deaths in the United States during the past year showed an average life span of 71.8 years. Assuming a standard deviation of 8.9 years, can we conclude that the actual life span is not 75 years? [pic]

7. A manufacturer of sports equipment has developed a new synthetic fishing line that he claims has a mean breaking strength of 8 kilograms. Test the hypothesis that the mean is 8 kg against the alternative that it is less than 8 kg, if a random sample of 50 lines is tested and found to have a mean breaking strength of 7.8 kg. The standard deviation is 0.8 kg. Use a 0.01 level of significance.

...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....

...MAT 300: STATISTICS
M&M PROJECT PAPER
ALEXANDREA WINT
PROFESSOR AZAD, VARGHA
June 3, 2012
Purpose of Report
The purpose of this project is to find the information for a quality control manager of Masterfoods plant. The manager wants to know about the proportion of candies and if they are the same or different. If there is any difference that exists then the manager wants to know why there is a difference in such cases. A study was conducted and results were obtained and based on these statistical results we will try to gain information about the quality of Masterfoods plant.
Part1: Sampling Method
Sampling method is used to draw the random sample from the population of candies from different bags is known as the simple random sampling without replacement. 36 bags were drawn from the population of different color of candies and after drawing the 36 samples; the numbers of different candies are calculated. For sample in bag one how many are blue candies, how many are red candies, how many are yellow candies etc. There are six different colors in each bag. So the record is made after collecting the sample.
Part 2: Method, Analysis, Results
In this section we have made attempts to find the descriptive statistics related to the number of candies or proportion of candies in the bag. The mean number of candies per bag is approximately equaled to 55. This means that we are expecting 55 candies with a standard deviation of...

...
November 19, 2010
NAME: The Statistics of Poverty and Inequality
TYPE: Sample
SIZE: 97 observations, 8 variables
DESCRIPTIVE ABSTRACT:
For 97 countries in the world, data are given for birth rates, death
rates, infant death rates, life expectancies for males and females, and
Gross National Product.
SOURCES:
Day, A. (ed.) (1992), _The Annual Register 1992_, 234, London:
Longmans.
_U.N.E.S.C.O. 1990 Demographic Year Book_ (1990), New York: United
Nations.
VARIABLE DESCRIPTIONS:
Columns
1 - 6 Live birth rate per 1,000 of population
7 - 14 Death rate per 1,000 of population
15 - 22 Infant deaths per 1,000 of population under 1 year old
23 - 30 Life expectancy at birth for males
31 - 38 Life expectancy at birth for females
39 - 46 Gross National Product per capita in U.S. dollars
47 - 52 Country Group
1 = Eastern Europe
2 = South America and Mexico
3 = Western Europe, North America, Japan, Australia, New Zealand
4 = Middle East
5 = Asia
6 = Africa
53 - 74 Country
Values are aligned and delimited by blanks.
Missing values are denoted with *.
The Statistics of Poverty and Inequality
This paper describes a case study based on data taken from the U.N.E.S.C.O. 1990 Demographic Year Book and The Annual Register 1992 giving birth rates, death rates, life expectancies, and Gross National Products for 97 countries.
When reviewing the statistics...

...Key Synthesis/Potential Test Questions (PTQs)
• What is statistics? Making an inference about a population from a
sample.
• What is the logic that allows you to be 95% confident that the confidence interval contains the population parameter?
We know from the CLT that sample means are normally distributed around the real population mean (). Any time you have a sample mean within E (margin of error) of then the confidence interval will contain . Since 95% of the sample means are within E of then 95% of the confidence interval constructed in this way will contain.
• Why do we use confidence intervals verses point estimates? The sample mean is a point estimate (single number estimate) of the population mean – Due to sampling error, we know this is off. Instead, we construct an interval estimate, which takes into account the standard deviation, and sample size.
– Usually stated as (point estimate) ± (margin of error)
• What is meant by a 95% confidence interval? That we are 95% confident that our calculated confidence interval actually contains the true mean.
• What is the logic of a hypothesis test?
“If our sample result is very unlikely under the assumption of the null hypothesis, then the null hypothesis assumption is probably false. Thus, we reject the null hypothesis and infer the alternative hypothesis.”
• What is the logic of using a CI to do a HT?
We are 95% confident the proportion is in this interval… if the sample mean or...

...
Investigating Bottling Company Case Study
T.P
University
Statistics
Mat 300
Mr. Thevar
December 01, 2013
Investigating Bottling Company Case Study
The case study that is being investigated is for a bottling company producing less soda than what is advertised. Customers have complained that the sodas in the bottles contain less than the advertised sixteen ounces. The employees at the company have measured the amount of soda contained in each bottle. There are thirty bottles that have been pulled from the shelves. The manager of the company would like to have a detailed report on the possible causes, if any, for the shortage in the amount of soda or if the claim is not supported explain how to mitigate the issue in the future. In order to statistically find a cause in the shortage a hypothesis testing is conducted by finding the mean, median, and standard deviation for ounces in the bottles. Constructing a 95 percent interval will establish the mean of the population since the mean of the population is not known.
There are thirty soda bottles being pulled for investigation. The mean will be calculated by averaging the amount of ounces in each bottle and dividing the total by the number of bottles. The data below shows the ounces in each of the thirty bottles that were pulled.
The mean among the sample bottles is 14.87. The calculation to find the mean is to add all the ounces per bottle. The total is 446.1 divided...

...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...