Chapter 7, Problem 7.1.1RE

Eye Blinks A survey of 49 individuals found that the average number of times a person blinks his or her eyes is 25 times per minute. Based on this survey, find the 90% confidence of the population mean number of eye blinks per minute. The population standard deviation is 4.

To determine

To construct: The 90% confidence interval for the populationmean number of eye blinks per minute.

Answer to Problem 7.1.1RE

The 90% confidence interval for the population mean number of eye blinks per minuteis $\underset{\_}{24.060<\mu <25.940}$.

Explanation of Solution

Given info:

$\overline{x}=25$, $\sigma =4$, and $n=49$.

Calculation:

Confidence interval:

Software Procedure:

Step by step procedure to obtain the 90% confidence interval for the mean bill for all cars from the drive-thruusing the MINITAB software:

• Choose Stat > Basic Statistics > 1-Sample Z.
• In Summarized data, enter the sample size as 49 and mean as 25.
• In Standard deviation, enter a value for s as 4.
• Check Options, enter Confidence level as 90.0.
• Choose not equal in alternative.
• Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

From the MINITAB output, the 90% confidence interval for the population mean number of eye blinks per minuteis $\underset{\_}{24.060<\mu <25.940}$