Question

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.

Answer 1

Answer:

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 $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.005 = 0.995$, so $z = 2.575$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

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

$M = 2.575\frac{1.4}{\sqrt{1233}} = 0.1$

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.

Answer 2

Answer: = ( 7.3, 7.5)

Therefore at 99% confidence interval (a,b) = ( 7.3, 7.5)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean gain x = 7.4

Standard deviation r = 1.4

Number of samples n = 1233

Confidence interval = 99%

z(at 99% confidence) = 2.58

Substituting the values we have;

7.4+/-2.58(1.4/√1233)

7.4+/-2.58(0.03987)

7.4+/-0.1028

7.4+/-0.1

= ( 7.3, 7.5)

Therefore at 99% confidence interval (a,b) = ( 7.3, 7.5)