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Point Estimate
| Point Estimate for p | ||||||||||||||||||||||||||||||
| x | 124 | |||||||||||||||||||||||||||||
| n | 427 | |||||||||||||||||||||||||||||
| p_hat | 0.2903981265 | |||||||||||||||||||||||||||||
| Point Estimate for mean | ||||||||||||||||||||||||||||||
| 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4.5 | 4.5 | 4.5 | 4.5 | 4.5 | 4.5 | 4.5 | 4.5 | 4.5 | 4.5 | 4.5 | 6 | 6 | 7 | 8 | 9 | 9 | 10 | 10 | 13 | 13 | 15 |
| n | 31 | |||||||||||||||||||||||||||||
| mean | 6.1774193548 | AVERAGE() | ||||||||||||||||||||||||||||
| 6.1774193548 |
Example 1: A recent survey of 1400 airline passengers suggests that 38% of air passengers prefer aisle seats. Solution: Since the sample proportion is the best point estimate of the population proportion, we conclude that the best point estimate of p is 0.38 . You'll notice that 0.38 represents the 38% of airline passengers which prefer aisle seats. Example 2: A Consumer Reports Research Center survey of 427 women showed 124 of them purchase books online. Find the best point estimate of the population proportion of women who purchase books online. Solution: x = the number of women who purchase books online n = total number of women in the survey The best point estimate of the population proportion of women who purchase books online is 0.290.
Example 2 (corrected): A random sample of college students were survey to find the numbers of years it took them to earn bachelor's degrees. Find the point estimate of the mean time required for all college students to earn bachelor's degrees.
Critical Values
| Finding Critical Values | |||||
| CL | alpha | area | critical value | Function | |
| If CL = 95%, z = 1.96 | 0.95 | 0.05 | 0.975 | 1.96 | NORM.S.INV() |
| If CL = 90%, z = 1.645 | 0.9 | 0.1 | 0.95 | 1.645 | |
| If CL = 99%, z = 2.58 | 0.99 | 0.01 | 0.995 | 2.58 | |
| Assuming n = 40 | CL | alpha | area | critical value | Function |
| If CL = 95%, t = 2.02 | 0.95 | 0.05 | 0.975 | 2.02269092 | T.INV() |
| If CL = 90%, t = 1.685 | 0.9 | 0.1 | 0.95 | 1.6848751217 | |
| If CL = 99%, t = 2.71 | 0.99 | 0.01 | 0.995 | 2.7079131835 | |
| n | 40 | ||||
| d.f. | 39 |
Note: Typically, for t-distribution, you need n > 30. If the population appears normally distributed, then n can be any number.
E and Confidence Interval
| Margin of Error in Real Life | ||||||||||||||||||||||||||||||
| Margin of Error, Confidence Interval for p | ||||||||||||||||||||||||||||||
| p-hat | 0.7 | |||||||||||||||||||||||||||||
| q-hat | 0.3 | |||||||||||||||||||||||||||||
| CL | 0.95 | |||||||||||||||||||||||||||||
| area | 0.975 | |||||||||||||||||||||||||||||
| If CL = 95%, z = 1.96 | 1.96 | |||||||||||||||||||||||||||||
| n | 1501 | |||||||||||||||||||||||||||||
| Margin of Error (E) | 0.0231828803 | |||||||||||||||||||||||||||||
| or | ||||||||||||||||||||||||||||||
| 2.32% | ||||||||||||||||||||||||||||||
| Low end | 0.6768171197 | |||||||||||||||||||||||||||||
| High end | 0.7231828803 | |||||||||||||||||||||||||||||
| FYI: | ||||||||||||||||||||||||||||||
| If CL = 95%, z = 1.96 | 0.975 | 1.96 | ||||||||||||||||||||||||||||
| If CL = 90%, z = 1.645 | 0.95 | 1.645 | ||||||||||||||||||||||||||||
| If CL = 99%, z = 2.58 | 0.995 | 2.58 | ||||||||||||||||||||||||||||
| Margin of Error, Confidence Interval for mean | ||||||||||||||||||||||||||||||
| 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4.5 | 4.5 | 4.5 | 4.5 | 4.5 | 4.5 | 4.5 | 4.5 | 4.5 | 4.5 | 4.5 | 6 | 6 | 7 | 8 | 9 | 9 | 10 | 10 | 13 | 13 | 15 |
| n | 31 | |||||||||||||||||||||||||||||
| mean (x_bar) | 6.1774193548 | AVERAGE() | ||||||||||||||||||||||||||||
| standard dev (s) | 3.1292927505 | STDEV.S() | ||||||||||||||||||||||||||||
| d.f. | 30 | n-1 | We use Student's t distribution, because 1) we don't know sigma, and 2) n > 30. | |||||||||||||||||||||||||||
| CL | 0.95 | |||||||||||||||||||||||||||||
| area | 0.975 | |||||||||||||||||||||||||||||
| t | 2.0422724563 | |||||||||||||||||||||||||||||
| Margin of Error | 1.1478338478 | Use SQRT() | 1.1478338478 | |||||||||||||||||||||||||||
| Low end | 5.0295855071 | |||||||||||||||||||||||||||||
| High end | 7.3252532026 |
A survey found that 70% of 1501 randomly selected responded favors a tax relief. Find the margin of error and the confidence interval given that the confidence level is 95%.
The margin of error is about 2.3%.
Example 2 (corrected): A random sample of college students were survey to find the numbers of years it took them to earn bachelor's degrees. Construct a 95% confidence interval estimate of the mean time required for all college students to earn bachelor's degrees.
The margin of error is about 1.14.
Sample Size
| Sample Size for p | |||
| p_hat | 0.84 | ||
| q_hat | 0.16 | ||
| If CL = 90%, z = 1.645 | 0.95 | 1.645 | |
| E | 0.03 | ||
| n | 404.0278224782 | 404 | ROUND() |
| Sample Size for mean | |||
| z | 1.96 | ||
| Std dev | 15 | ||
| E | 5 | ||
| n | 34.5744 |
Suppose at the confidence level of 95%, we want to calculate the necessary sample size for σ = 15. Assuming that we know E = 5, calculate n.
A sample of about 35 is needed.
Example 6: You have been given a task of estimating the percentage of Southwest flights that arrive on time, which is no later than 15 minutes after the scheduled arrival time. How many flights must you survey to be 90% confident that your estimate is within three percentage points of the true population percentage? Assume that for a recent year, 84% of Southwest flights were on time.
Recommended reading prior to starting this assignment: Chapter 6.4 & 6.5
·
· Given a confidence level C, determine the critical value (z/2) from the standard normal table or Excel needed to construct a confidence interval for p.
·
·
· Calculate the sample size needed to estimate a population proportion p.
·
· Verify the requirements needed to estimate a population mean
· Explain the purpose of the t-distribution and when it is used.
· Determine the appropriate degrees of freedom associated with t distribution.
· Given a confidence level C, determine the t critical value (t/2) from the t distribution table or Excel
·
·
· Calculate the sample size needed to estimate a population proportion .
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Recommended reading prior to starting this assignment: Chapter 6.4 & 6.5
Given a study, find the best point estimate
ˆp
of the population proportion p.
Given a confidence level C, determine the critical value ( z
/2
) from the standard normal table or Excel needed to construct a
confidence interval for p.
Find the margin of error E needed to construct a confidence interval for p using the formula
/2ˆˆpqEzn
.
Construct and interpret the confidence interval for a population proportion p using the formula
ˆˆpEppE
.
Calculate the sample size needed to estimate a population proportion p.
Given a study, find the best point estimate
x
of the population mean
.
Verify the requirements needed to estimate a population mean
Explain the purpose of the t-distribution and when it is used.
Determine the appropriate degrees of freedom associated with t distribution.
Given a confidence level C, determine the t critical value (t
/2
) from the t distribution table or Excel
needed to construct a confidence interval for
.
Find the margin of error E needed to construct a confidence interval for
using the formula
/2sEtn
.
Construct and interpret the confidence interval for a population proportion using the formula
xExE
.
Calculate the sample size needed to estimate a population proportion .
