One Sided Confidence Interval & Estimation of Sample Size Lecture 27

 One-Sided Confidence Interval & Estimation of Sample Size

Lecture 27

One-Sided Confidence Interval

The lower and upper limits of a normal population's mean are sometimes sought after. For instance, a businessman could want to know the deal's minimum or maximum profit. After that, a one-sided confidence interval is employed. In such cases, a full significance level in a one-sided sampling distribution and the marginal error become zα ×SE.

For lower limit 

For upper limit

Example 7.8: A zoologist would want to know the average weight of a certain species of Australian diamond finch. A zoologist might instead gather 50 diamond finches and measure the sample's mean weight (30 ounces) and standard deviation (0.13 ounces), as it would take too long to weigh thousands of individual finches. Determine the 95% confidence level for the minimal weight estimate.

Solution: n = 50, sample mean = 30 standard deviation = 0.13

1- α = 0.95

 α = 0.05

z α = z0.05 = 1.645

The 95% confidence limit (minimal weight estimate) is given by:

Example 7.9: A grocery company's marketing staff might execute two distinct advertising campaigns at 20 different locations each during a quarter. At the conclusion of the quarter, the average sales at each store would be $2,000 with a standard deviation of $36. Calculate the highest profit with 98% accuracy.

 

Solution: n = 20, sample mean = $ 2000, standard deviation = $ 36.

 

1- α = 0.98

 α = 0.02

z α = z0.02 = 2.33

Estimation of Sample Size

Estimation of the sample size is an important step in the planning and estimating the time and cost needed to carry out the research study. The estimation of sample size depends on the following:

1. Standard deviation

2. The confidence coefficient

3. Error of the estimate (the difference between estimate and true value).

4. Population size 

The Yamane Formula, a commonly used mathematical formula for determination of sample size, is given by:

In case the sample is selected by replacement or the sample size is less than 5% of the population size.

 

In case the sample is selected without replacement and the sample size is at least 5% of the population size.

Example 7.9: Finding the mean of a normal population requires a sample size large enough that there is a 0.95 probability that the sample mean will not differ from the true mean by more than 25% of the standard deviation. What is the appropriate sample size?

Solution:

1-α =0.95

α = 0.05

Z0.05/2 = 1.96

e = 25 % of σ

e = 25/100 σ

e = σ / 4

Example 7.10: If the standard deviation of the normal distribution is 10 and the estimate error is 1.52 with a 95% confidence level, what number of samples should be used?

Solution:

1-α =0.95

α = 0.05

Z0.05/2 = 1.96

e = 1.50

σ = 10

Example 7.11: A population consists of 2000 observations with a standard deviation of 2.10. and a probable error of 1.60 with a 98% confidence coefficient. What should be the sample size that will be taken without replacement? 

Solution:

N = 2000

1-α =0.98

α = 0.02

Z0.02/2 = 2.33

e = 1.60

σ = 2.10

The sample is selected without replacement:

Confidence Interval for Population Proportion

Let p^ be the sample proportion computed from the values of a random sample of size n selected from a binomial population, with the proportion of success being P. If the sample size is large enough, then the sampling of p^ approach to normal distribution with mean = 0 and standard deviation is 1.

Now to construct a confidence interval for P, choose two values from the z table and make the following probability statement.

Example 7.12: A random sample of size 625 is selected from a binomial distribution with a proportion of success of 0.62. Determine a 90% confidence interval for the proportion of success of the entire population.

Solution:

n = 625,  p^ = 0.62

 q^ = 1 -  p^

 q^ = 1 -  0.62

 q^ = 0.38

1-α= 0.90

α =0.02

Z0.02/ 2 = 1.645

Hence, 95% confidence limits for the proportion of success are 0.588 to 0.652.

Example 7.13: A random sample of 500 individuals surveyed, 320 said they are graduates. Find the true percentage of individuals in this city who are graduates using a confidence interval estimate with a 95% confidence level.

Solution: Let p^ is the proposition of graduate in a sample.

p^ = Number of graduates / Sample size

p^ = 320 / 500

p^ = 0.64

q^ = 1 - p^

q^ = 1- 0.64

q^ = 0.36

1-α =0.95

α = 0.05

Z0.05/2 = 1.96

Estimation of Sample Size

The number of observations gathered during a study is referred to as the sample size, and it is a crucial factor in determining the precision and dependability of research conclusions. A well-calculated sample size minimizes bias while optimizing precision by guaranteeing that the findings are representative of the population.

The determination of sample size depends on the following factors:

i. Proportion of success.

ii. confidence coefficient.

iii. Error of the estimate.

Example 7.14: 10 of the 600 randomly selected units of 30-watt energy saver bulbs are defective, in accordance with the company’s specification. How large a sample is required if we want 95% confidence that the error of the estimate is less than 0.05?

solution:

p^ = Number of defectives / Sample size

p^ = 10 / 600

p^ = 0.017

q^ = 1 - p^

q^ = 1- 0.017

q^ = 0.983

1-α =0.95

α = 0.05

Z0.05/2 = 1.96

e = 0.05

• Read More: Hypothesis Testing