A sample of n = 15 scores has a mean of M = 8. Another sample of n = 5 scores has a mean of M = 4. If these two samples are combined, what is the mean for the combined group? Note that since n is different for each sample of scores, you cannot simply calculate the average of the two means. Youmust calculate the weighted mean.

I'm assuming here that 1,040 means one thousand and forty, not a metric decimal comma.

First we need the Std Dev of the mean value itself. This is

s / sqrt (n) = 21000 / sqrt (1040) = 651.2

Then the confidence interval for the mean value is

mean value ± Z * ( s / sqrt (n) )

where Z is an estimator related to the required confide

CL Z
99% 2.576
98% 2.326
95% 1.96
90% 1.645

So for a confidence of 95% the margin of error for the true value of the mean is

46239 ± 1.96 * 651.2 or
46239 ± 1276.3

that is, the mean of a sample from this process will be in this range to 95% confidence.

The margin of error for a future sample value would be

46239 ± 1.96 * 21000 or

46239 ± 41160

that is, a single sample value from this process will be in this range to 95% confidence.
.

First you need the Std Error of the mean value, a measure of the dispersion of that mean value.

SE = sample std deviation / sqrt (sample size)
= 100 / sqrt (64)
= 100 / 8
= 12.5

Then we use a figure for the number of std errors either side of the mean value, which make up a 95 % confidence interval. This is ± 1.96 std errors, from tables of the Normal Distribution.

So the confidence interval is

350 ± 1.96 * 12.5 or

374.5 to 325.5
.

First you need the Std Error of the mean value, a measure of the dispersion of that mean value.

SE = sample std deviation / sqrt (sample size)
= 8 / ? 64
= 1

Then we use a figure for the number of std errors either side of the mean value, which make up a 99 % confidence interval. This is ± 2.58 std errors, from tables of the Normal Distribution.

So the confidence interval is

125 ± 2.58 * 1 or

127.58 to 122.42
.

Oh geeze,, you have 480 ladybugs in all?!!!
That makes a total combined population of 1,680 critters!
And why did you only take a 1.25% sample?!!

But maybe ladybugs like living in dirt piles better than other backyard areas, so you might not have that many! ;)

No. At the 0.05 level you cannot conclude that this sample is significantly different.

Count the number of items in the sample whose statistics you're calculating. For example, if you're calculating the mean and standard deviation of seven items then the sample size is seven.

Subtract the mean and then divide by the standard deviation.
If you need further information, please reply to this post and specify the make and model of your calculator. You can also post a sample problem if you wish to see the solution.

For audio:

MP3 (8-320kbps and VBR; Sample rates: 8, 11.025, 16, 22.050, 24, 32, 44.1, 48kHz)
WMA (32-192kbps; Sample rates: 8, 11.025, 16, 22.050, 24, 32, 44.1, 48kHz)

For video:
WMV (30fps, 512kbps; 320 x 240 pixels)

Hope this helps.

check bottom of the laptop

1 Model Number dv9095xx (Sample A)
dv7-1000 (Sample B)
15-1001xx (Sample C) 2 Product Number RQ877AS (Sample A)
AA123AV (Sample B)
VR898AS (Sample C) 3 Product Name HP Pavilion dv9000 (Sample A)
Pavilion dv7 (Sample B)
HP ENVY 15 (Sample C)

add all the scores up and divide to the number of victories
so that's (32+12+28+22+20+25+22+19)/8 = 22.5