Caution!!! Unless n is extremely large, the large sample procedure performs poorly when p is near 0 or near 1. For instance ; p= 0.001 and n=100 so that np=0.1 <15 Unless we have an extremely large n we can not satisfy the sample size assumption. However, making minor adjustment on the sample proportion can handle that problem.

Adjusted (1 – )100% Confidence Interval for a Population Proportion, p the adjusted sample proportion of observations with the characteristic of interest, x is the number of successes in the sample, and n is the sample size.

Example According to the Bureau of Labor Statistics (2012), the probability of injury while working at a jewelry store is less than 0.01. Suppose that in a random sample of 200 jewelry store workers, 3 were injured on the job. Estimate the true proportion of jewelry store workers who are injured on the job using 95% confidence interval.

Example-Solution We are 95% confident that the true proportion of jewelry store workers who are injured while on the job falls between 0.004 and 0.046. If we had applied the standard large-sample confidence interval formula, we would have obtained the following interval which has negative lower bound.

Sample Size Determination for 100(1 – ) % Confidence Interval for p In order to estimate p with a fixed margin of error and with 100(1 – )% confidence, the required sample size is found by solving the following equation for n: When true proportion is unknown, we use sample proportion from a prior sample instead.

The value of p(1-p) has its maximum when p=0.5 or values close to 0.5. So, we can obtain conservatively large values of n by approximating p by 0.5 or values close to 0.5. In any case we should round the value of n obtained upward to ensure that the sample size will be sufficient to achive specified reliability.

Example A cellular telephone manufacturer that entered the postregulation market too quickly has an initial problem with excessive customer complaints and consequent returns of the cell phones for repair or replacement. The manufacturer wants to determine the magnitude of the problem in order to estimate its warranty liability. How many cellular telephones should the company randomly sample from its warehouse and check in order to estimate the fraction defective, p, to within 0.01 with 90% confidence? Assume proportion of number of defective cellular phones is 0.10.

Example-solution

Confidence Interval for a Population Mean: Normal (z) Statistic 6.2 Confidence Interval for a Population Mean: Normal (z) Statistic :1, 1, 3

Estimation Process Population Random Sample              I am 95% confident that  is between 40 & 60. Mean x = 50 Mean, , is unknown     Sample         7

Confidence Interval According to the Central Limit Theorem, the sampling distribution of the sample mean is approximately normal for large samples. Let us calculate the interval estimator: That is, we form an interval from 1.96 standard deviations below the sample mean to 1.96 standard deviations above the mean. Prior to drawing the sample, what are the chances that this interval will enclose µ, the population mean?

Confidence Interval If sample measurements yield a value of that falls between the two lines on either side of µ, then the interval will contain µ. The area under the normal curve between these two boundaries is exactly .95. Thus, the probability that a randomly selected interval will contain µ is equal to .95.

95% Confidence Level If our confidence level is 95%, then in the long run, 95% of our confidence intervals will contain µ and 5% will not. For a confidence coefficient of 95%, the area in the two tails is .05. To choose a different confidence level we increase or decrease the area (call it ) assigned to the tails. If we place /2 in each tail and z/2 is the z-value, the confidence interval with coefficient (1 – ) is

Conditions Required for a Valid Large-Sample Confidence Interval for µ 1. A random sample is selected from the target population. 2. The sample size n is large (i.e., n ≥ 30). Due to the Central Limit Theorem, this condition guarantees that the sampling distribution of is approximately normal. Also, for large n, s will be a good estimator of .

Large-Sample (1 – )% Confidence Interval for µ where z/2 is the z-value with an area /2 to its right and in the standard normal distribution. The parameter  is the standard deviation of the sampled population, and n is the sample size. Note: When  is unknown and n is large (n ≥ 30), the confidence interval is approximately equal to where s is the sample standard deviation.

Thinking Challenge You’re a Q/C inspector for Gallo. The  for 2-liter bottles is .05 liters. A random sample of 100 bottles showed x = 1.99 liters. What is the 90% confidence interval estimate of the true mean amount in 2-liter bottles? 2 liter 2 liter © 1984-1994 T/Maker Co.

Confidence Interval Solution* We can be 90% confident that the mean amount in 2-liter bottles between 1.982 and 1.998. Our confidence is derived from the fact that 90% of the intervals formed in repeated applications of this procedure would contain  52

Exercise A random sample of 70 observations from a normally distributed population possesses a sample mean equal to 26.2 and a sample standard deviation equal to 4.1 A) Find an approximate 95% confidence interval for . B) What do you mean when you say that a confidence level is 95%? C) Find an approximate 99% confidence interval for . D) What happens to the width of a confidence interval as the value of the confidence coefficient is increased while the sample size is held fixed?

Thinking Challenge We have a random sample of customer order totals with an average of $78.25 and a population standard deviation of $22.5. A) Calculate a 90% confidence interval for the mean given a sample size of 40 orders. B) Calculate a 90% confidence interval for the mean given a sample size of 75 orders. C) Explain the difference in the 90% confidence intervals calculated in A and B. Calculate the minimum sample size needed to identify a 90% confidence interval for the mean assuming a $5 margin of error.