Suppose 100 randomly selected customers were asked to rate a particular service they recently had on a scale of 0 to 10. The sample mean rating was found to be 7.3. (Assume the population standard deviation of all ratings in the population of customers is 0.8.) Using all this information, estimate the average rating of this service for all customers.

What level of significance are you using for your estimate? If P = .05:

95% = mean ± 1.645 SEm

SEm = SD/√n

To estimate the average rating of the service for all customers using the given information, we can use the concept of confidence intervals.

First, it's important to note that the sample mean rating of 7.3 is an estimate of the population mean rating. The goal is to find a range around this estimate that is likely to contain the true population mean rating.

To do this, we can use the formula for a confidence interval. The formula is:

Confidence Interval = Sample Mean ± (Z * Standard Error)

In this case, we know the sample mean rating is 7.3. The Z-value represents the number of standard deviations to use to create the interval and depends on the desired confidence level. Assuming a 95% confidence level, the Z-value is approximately 1.96 (which corresponds to a 1.96 standard deviation).

The standard error is the estimated standard deviation of the sample mean rating, and it can be calculated using the formula:

Standard Error = Population Standard Deviation / √(Sample Size)

Given that the population standard deviation is 0.8 and the sample size is 100, we can calculate the standard error:

Standard Error = 0.8 / √100 = 0.8 / 10 = 0.08

Now, we can plug in the values into the confidence interval formula:

Confidence Interval = 7.3 ± (1.96 * 0.08)

Calculating this, we get:

Confidence Interval = 7.3 ± 0.1568

Therefore, the estimated average rating of the service for all customers is within the range of 7.1432 to 7.4568 (rounding to four decimal places).

In other words, we can be 95% confident that the true average rating of the service lies between 7.1432 and 7.4568 based on the given sample data.