Statistics Construction of Confidence Intervals

Construct Confidence Intervals
Part 1
1. Explain the difference between a 95% confidence interval and a 99% confidence interval in terms of probability
a) To construct a 95% confidence interval for a population, mean µ, what is the correct critical value z*?
A 95% confidence interval implies that in the event that 100 different kinds of samples are taken into consideration and a 95% confidence interval is calculated for every sample, then roughly 95 out of the 100 confidence intervals will have the true mean value, which is µ
To construct a 95% confidence interval for a population, mean µ, the correct critical value of z* is P (-1.96 < Z < 1.96) = 0.95
b) To construct a 99% confidence interval for a population, mean µ, what is the correct critical value z*?
A 99% confidence interval implies that in the event that 100 different kinds of samples are taken into consideration and a 99% confidence interval is calculated for every sample, then roughly 99 out of the 100 confidence intervals will have the true mean value, which is µ
To construct a 99% confidence interval for a population, mean µ, the correct critical value of z* is P (-2.56 < Z < 2.56) = 0.99
2. Explain what the margin of error is and how to calculate it
Margin of error is defined as the range of values both above and below the sample statistics within a confidence interval. Therefore, the margin of error indicates the number of percentage points the outcomes will vary from the real value of the population.
The margin of error can be computed in two ways:
1. Margin of error = Critical value * Standard deviation
2. Margin of error = Critical value * Standard error of the statistic (Mendenhall, Beaver, & Beaver, 2012)
3. A survey of a group of students at a certain college, we call College ABC, asked: “About how many hours do you study in a week?” The mean response of the 400 students is 15.8 hours. Suppose that the study time distribution of the population is known to be normal with a standard deviation of 8.5 hours. Use the survey results to construct a 95% confidence interval for the mean study time at the College ABC.
In this case:
Population Mean X = 15.8
N = 400
Standard Deviation ? = 8.5
Therefore, 95% confidence interval for the population mean is computed as follows:
X ± 1.96 ? / ?N
= 15.8 ± 1.96 8.5 / ?400
= 15.8 ± 0.833
= (14.967, 16.633)
4. Explain the difference between a null hypothesis and an alternative hypothesis
The null hypothesis is considered to be the hypothesis of no variance or no change. The researcher makes the assumption that claim is true up until sample results specify otherwise. On the other hand, alternative hypothesis is the hypothesis that the researcher wishes to prove or authenticate. That is a statement regarding the value of a parameter that is either of a lesser amount than, bigger than, or not equal to the claimed parameter.
5. Suppose that you are testing a null hypothesis H0: µ = 10 against the alternate H1: µ ? 10. A simple random sample of 35 observations from a normal population are used for a test. What values of the z statistic are statistically significant at the ? = 0.05 level?
X ± 1.96 ? / ?N
= 10 ± 1.96 ? / ?35
6. Describe the four-step process for tests of significance according to the textbook
The main objective of hypothesis testing is to make a determination of the probability that a population factor, for instance, the mean of the population, can be true. The first step of hypothesis testing is stating the hypotheses, that is, the null hypothesis and the alternative hypothesis. The second step is setting the criteria for making a decision. The third step is computing the test statistic. The final step is making a decision.
Part 2
1. A study of a group of 40 male league bowlers chosen at random had an average score was 176. It is known that the standard deviation of the population is 9.
a) Construct the 95% confidence interval for the mean score of all league bowlers.
The 95% confidence interval for the population mean is computed as follows:
X ± 1.96 ? / ?N
= 176 ± 1.96 (9 / ?40)
= 176 ± 2.79
= (173.21, 178.79)
b) Construct the 95% confidence interval for the mean score of all league bowlers assuming that a sample of size 100 is used instead of 40, and the same mean and standard deviation occur.
The 95% confidence interval for the population mean is computed as follows:
X ± 1.96 ? / ?N
= 176 ± 1.96 (9 / ?100)
= 176 ± 1.764
= (177.764, 174.236)
c) Give the margin of error for each interval
Margin of error = Critical value * Standard deviation
The margin of error for part a is 2.79
The margin of error for part b is 1.764
d) Explain why one confidence interval is larger than the other.
The main reason why one confidence interval is larger than the other is because as the accuracy of the confidence interval increases then the reliability of an interval comprising of the actual mean decreases.
2. There are 100 apartments in a certain a San Francisco apartment building. The owner of the building wants to estimate the mean number of people living in an apartment. The owner draws a random sample of 40 apartments in the building. The number of people living in each apartment is as follows:
1 2 1 2 3 1 3 4 3 1
2 2 1 2 2 2 1 3 2 3
2 3 1 2 3 3 2 4 5 2
3 2 2 3 1 1 2 2 1 2
a) Compute the sample mean and sample standard deviation.
Sample mean = (? xi) / n
= 87 / 40
= 2.175
Sample standard deviation
Sample variance = 34.3525
Sample Standard Deviation = ?34.3525 = 5.861
b) Use the results from part (a) to construct a 95% confidence interval.
The 95% confidence interval for the population mean is computed as follows:
X ± 1.96 ? / ?N
= 2.175 ± 1.96 (5.861 / ?40)
= 2.175 ± 1.816
= (0.359, 3.991)
3. A doctor wishes to estimate the birth weights of infants. How large a sample must the doctor select if she desires to be 99% confident that the true population mean is at most 6 ounces away from the mean of the sample? Assume the standard deviation is 8 ounces. Hint: The margin of error should be at most 6.
The 99% confidence interval for the population mean is 2.575
N = (2.575 (8) / 2)2
= 106.09
= 107
4. In Problem 4 of Part 1 a class survey of 400 students was given in which students at College ABC claimed to study an average of 15.8 hours per week. Consider these students as a simple random sample from the particular population of College ABC students. We want to investigate the question: Does the survey provide good evidence that students study more than 15 hours per week on average? Assume the population of hours studied is normal with a standard deviation of 4.
Therefore, 95% confidence interval for the population mean is computed as follows:
X ± 1.96 ? / ?N
= 15.8 ± 1.96 (4 / ?400)
= 15.8 ± 0.392
= (15.408, 16.192)
This shows that students study more than 15 hours per week on average
5. Before working out this problem, it will help to look over the webpage, Hypothesis tests for means, http://stattrek.com/hypothesis-test/mean.aspx?Tutorial=AP
a) State the null and alternate hypothesis in terms of the mean study time in hours for the population.
H0 = students study more than 15 hours per week on average
H1 = students study less than 15 hours per week on average
b) Is this a one-tailed test or two-tailed test?
This is a one-tailed test
c) Determine the value of the test statistic.
The test statistic is 0.392
d) Sketch a normal shape curve and identify the test statistic.


Test statistic


e) Indicate the p-value of the test. Use the standard normal table. Shade the area under the normal curve corresponding to the p-value. You can also use the website cited above to do this.
The P-Value is 0.347529.
f) State your conclusion to the statistical problem in terms of the null hypothesis, and your conclusion to the practical problem.
The null hypothesis is true










References
Mendenhall, W., Beaver, R. J., & Beaver, B. M. (2012). Introduction to probability and statistics. New York: Cengage Learning.