Question

A random sample of 64 door-to-door encyclopedia salespersons were asked how long on average they were able to talk to the potential customer. Their answers revealed a mean of 8.5 minutes. The population standard deviation is 3 minutes.

Construct a 95% confidence interval for mu, the time it takes an encyclopedia salesperson to talk to a potential customer.
What is the upper confidence limit?

Answer 1

The 95% confidence interval for mu, the time it takes an encyclopedia salesperson to talk to a potential customer is between 7.765 minutes and 9.235 minutes.

The upper confidence limit is of 9.12 minutes.

Explanation:

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

`\alpha = (1 - 0.95)/(2) = 0.025`

Now, we have to find z in the Z-table as such z has a p-value of
`1 - \alpha`.

That is z with a pvalue of
`1 - 0.025 = 0.975`, so Z = 1.96.

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

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

`M = 1.96(3)/(√(64)) = 0.735`

The lower end of the interval is the sample mean subtracted by M. So it is 8.5 - 0.735 = 7.765 minutes

The upper end of the interval is the sample mean added to M. So it is 8.5 + 0.735 = 9.235 minutes

The 95% confidence interval for mu, the time it takes an encyclopedia salesperson to talk to a potential customer is between 7.765 minutes and 9.235 minutes.

What is the upper confidence limit?

Similar procedue above, just a few changes.

Now Z with a p-value of 0.95, so Z = 1.645.

`M = 1.645(3)/(√(64)) = 0.62`

8.5 + 0.62 = 9.12 minutes

The upper confidence limit is of 9.12 minutes.