Question

1. Pizza for carry-out A sample of 300 orders for take-out food at a local pizzeria found that the average cost of an order was $23. Assume for now that the standard deviation of the population is indeed known and equals $15. Round final answers to 2 decimal places. (a) For a 99% confidence interval find the margin of error for the average cost of an order. (b) Find the 99% confidence interval. (c) Provide an interpretation of the confidence interval. (d) If we only need to be 95% confident, does the confidence interval become wider or narrower

Answer 1

a)
`ME = 2.58 (15)/(√(300))=2.234`

b)
`23-2.234=20.766`

`23+2.234=25.234`

c) We are 99% confident that the true mean for the costs of an order for the local pizzeria is between 20.766 and 25.234

d) For this case if we decrease the confidence then the margin of error would be lower and that implies a narrower interval. Becuase the critical value is lower if the decrease the confidence level

Explanation:

Data provided

`\bar X=23` represent the sample mean for the cost of orders

`\sigma =15` represent the sample population deviation

n=300 represent the sample size

Part a

The confidence interval for a true mean is given by:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

And the margin of error is given by:

`ME=t_(\alpha/2)(s)/(√(n))`

The Confidence is 0.99 or 99%, the significance is
`\alpha=0.01` and
`\alpha/2 =0.005`, and the critical value would be
`t_(\alpha/2)=2.58`

`ME = 2.58 (15)/(√(300))=2.234`

Part b

The confidence interval would be:

`23-2.234=20.766`

`23+2.234=25.234`

Part c

We are 99% confident that the true mean for the costs of an order for the local pizzeria is between 20.766 and 25.234

Part d

For this case if we decrease the confidence then the margin of error would be lower and that implies a narrower interval. Becuase the critical value is lower if the decrease the confidence level