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INTERVAL ESTIMATION. POPULATION MEAN: SIGMA ( ) KNOWN. You can find the problem we are going to solve here on page 306 of your textbook. 5. In an effort to estimate the mean amount spent per customer for dinner at a major Atlanta restaurant, data were collected for a sample of 49 customers. Assume a population standard deviation of $5. a. b. At 95% confidence, what is the margin of error? If the sample mean is $24.80, what is the 95% confidence interval for the population mean? Solution. a. We know that the margin of error is found using the following formula: Z 2 n Thus, in order to find the margin of error we need to find Z first. Since we want a confidence level of 2 95% this implies that Finding = 5%, which means that 2 Z 0.025 using a TI-83. Step 1. Press 2ND and VARS. Step 2. Go to 3: invNorm( and then press ENTER. = 0.025. Step3. Type 0.025) and press ENTER. Therefore, the margin of eror is equal to b. Now, using the fact that the sample mean is $24.8, we are going to to find a 95% confidence interval for . Step 1. Press STAT Step 2. Go to TESTS Step 3. Go to 7: Zinterval and press ENTER Step 4. Highlight Stats and press ENTER Step 5. Plug in the values of , x , n and the confidence level. Step 6. Highlight Calculate and press ENTER. You should see something like this. We are done!!