Question

The number of candy pieces in a bag is calculated for a sample of 40 bags of a particular manufacture. Based on this data, the sample mean is 28.2 pieces. The population standard deviation is to be 3.9 (remember this is the unrealistic part, we would not really know this). Which below is a 95% confidence interval for the mean number of pieces of candy per bag? Note: 95% of all possible confidence intervals for a sample size of 30 will contain the population mean. Use 1.960 for a 95% CI and 2.576 for a 99% CI

(26.99,29.41)
(24.31,32.09)
(20.17,36.23)
(26.24,30.16)

Answer 1

@amistre64

Answer 2

@kropot72 Can you help walk me through this... I keep getting the wrong answer every time

Answer 3

@mtbender74 @SarahEZZMcK

Answer 4

First you need to find the standard deviation of the sample mean. This is given by:
\[\large \sigma _{\bar{x}}=\frac{\sigma}{\sqrt{n}}\ .......(1)\]
So for this step, just plug the given values into (1).

Answer 5

The next step is to multiply the standard deviation of the sample mean by the the given value of z to find the 95% confidence interval. The result is called the 'margin of error'.
\[\large margin\ of\ error=1.96\times \sigma _{\bar{x}}=you\ can\ calculate\]

Answer 6

@bookworm1156 Are you there?

Answer 7

@bookworm1156 Can you calculate the standard deviation of the sample mean as follows:
\[\large \sigma _{\bar{x}}=\frac{3.9}{\sqrt{40}}=you\ can\ calculate\]

Answer 8

.6166441437

Answer 9

Then I take that number and multiply by 1.96 correct?

Answer 10

Correct. Now multiply your result by 1.96 to find the margin of error.

Answer 11

which would give me 1.208622522

Answer 12

Correct again. Now round that result for the margin of error to 1.21 and find the confidence interval as follows:
C.I. = (28.2 - 1.21), (28.2 + 1.21)

Answer 13

(26.99,29.41)

Quick Question? How do you find a t-score (or degrees of freedom for that matter?)

Answer 14

SO A is the correct answer!!

Answer 15

That is correct. Sorry, I don't have time right now to answer your other question.

Answer 16

that's ok. I can look it up. Thanks for your help!

Answer 17

You're welcome :)