A software developer wants to know how many new computer games people buy each year. A sample of 1233 people was taken to study their purchasing habits. Construct the 99% confidence interval for the mean number of computer games purchased each year if the sample mean was found to be 7.4. Assume that the population standard deviation is 1.4. Round your answers to one decimal place.

The 99% confidence interval for the mean number of computer games purchased each year is between 7.3 and 7.5 games.

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 7.4 - 0.1 = 7.3.

The upper end of the interval is the sample mean added to M. So it is 7.4 + 0.1 = 7.5

The 99% confidence interval for the mean number of computer games purchased each year is between 7.3 and 7.5 games.

what the question so we can you out

7 ft 6 in means 7 1/2, because 6 in is 1/2 of a foot. 7 ft 8 in means 7 2/3 because 8 in is 2/3 of a foot. 23/3 = 7 2/3. 15/2 = 7 1/2.

23/3   *   15/2   = 57 1/2

step-by-step explanation: