Suppose the degree of confidence is (1-a )´100%
Thus $502
2
1,
+ =
- n
S
x t
n
a
Here x = 487, S = 48, n = 100
2 2
99,
\t a » Za
Thus we get 502
10
48
487
2
+ = Za
Or 3.125
4.8
15
2
= = Za
0.0009 1 0.9982
2
\a = -a =
or
\ We can assert with 99.82% confidence that the true mean salaries will be between
$472 and $502.
It goes up.
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Did you mean, "How do you calculate the 99.9 % confidence interval to a parameter using the mean and the standard deviation?" ? The parameter is the population mean μ. Let xbar and s denote the sample mean and the sample standard deviation. The formula for a 99.9% confidence limit for μ is xbar - 3.08 s / √n and xbar + 3.08 s / √n where xbar is the sample mean, n the sample size and s the sample standard deviation. 3.08 comes from a Normal probability table.
The standard deviation of the population. the standard deviation of the population.
The standard deviation is 0.
The standard deviation is used in the numerator of the margin of error calculation. As the standard deviation increases, the margin of error increases; therefore the confidence interval width increases. So, the confidence interval gets wider.
no
The standard deviation associated with a statistic and its sampling distribution.
The increase in sample size will reduce the confidence interval. The increase in standard deviation will increase the confidence interval. The confidence interval is not based on a linear function so the overall effect will require some calculations based on the levels before and after these changes. It would depend on the relative rates at which the change in sample size and change in standard deviation occurred. If the sample size increased more quickly than then standard deviation, in some sense, then the size of the confidence interval would decrease. Conversely, if the standard deviation increased more quickly than the sample size, in some sense, then the size of the confidence interval would increase.
It goes up.
For normally distributed data. One standard deviation (1σ)Percentage within this confidence interval68.2689492% (68.3% )Percentage outside this confidence interval31.7310508% (31.7% )Ratio outside this confidence interval1 / 3.1514871 (1 / 3.15)
this dick
It will make it wider.
Did you mean, "How do you calculate the 99.9 % confidence interval to a parameter using the mean and the standard deviation?" ? The parameter is the population mean μ. Let xbar and s denote the sample mean and the sample standard deviation. The formula for a 99.9% confidence limit for μ is xbar - 3.08 s / √n and xbar + 3.08 s / √n where xbar is the sample mean, n the sample size and s the sample standard deviation. 3.08 comes from a Normal probability table.
decreases
It can be.
The confidence intervals will increase. How much it will increase depends on whether the underlying probability model is additive or multiplicative.