To obtain 95% confidence intervals for a normal distribution with known variance, you take the mean and add/subtract \(\displaystyle 1.96\times standard\ error\). This is because 95% of the values drawn from a normally distributed sampling distribution lie within 1.96 standard errors from the sample mean. 

Standard deviation for the samle mean:

\(\displaystyle 2500/\sqrt{24}=510.2\)

Since n < 30, we must use the t-table (not the z-table).

The 98% t-value for n=24 is 2.5.

\(\displaystyle 11,000\pm2.5\times510.2=11,000\pm1275.5\)

Because we have such a large sample size, we are using the standard normal or z-distribution to calculate the confidence interval.

Formula: 

\(\displaystyle \bar{x} \pm z \cdot \frac{\sigma}{\sqrt{n}}\)

\(\displaystyle \bar{x}= 0.978 g\) 

\(\displaystyle \sigma= 0.042\) 

We must find the appropriate z-value based on the given \(\displaystyle \alpha\) for 95% confidence: 

\(\displaystyle 1-\frac{\alpha}{2}= 1-\left(\frac{0.05}{2}\right)= 0.975\)

Then, find the associated z-score using the z-table for 

\(\displaystyle z _{0.975} = 1.96\)

Now we fill in the formula with our values from the problem to find the 95% CI.  

\(\displaystyle 0.978 \pm 1.96 \left( \frac{0.042}{\sqrt{300}}\right)= (0.973 g, 0.9827 g)\)

Because we are only given the sample standard deviation we will use the t-distribution to calculate the confidence interval.

Appropriate Formula: 

\(\displaystyle = \bar{x} \pm t \cdot \frac{s}{\sqrt{n}}\)

Now we must identify our variables:

\(\displaystyle \bar{x}=249\)

\(\displaystyle s=24\)

\(\displaystyle n=56\)

We must find the appropriate t-value based on the given \(\displaystyle \alpha\)

t-value at 90% confidence:

 \(\displaystyle t_{\frac{\alpha }{2}, n-1}= t _{\frac{0.10}{2}}= t_{0.05, 55}\)

Look up t-value for 0.05, 55 , so t-value= ~ 1.6735

90% CI becomes:

\(\displaystyle 249 \pm 1.6735\left(\frac{24}{\sqrt{56}}\right)\)

\(\displaystyle = (243.63 mm^3/g/hr, 254.36 mm^3/g/hr)\)

First you must calculate the sample mean and sample standard deviation of the sample.

\(\displaystyle \bar{x}=16.53\)

\(\displaystyle \sigma=5.29\)

Because we do not know the population standard deviation we will use the t-distribution to calculate the confidence intervals. We must use standard error in this formula because we are working with the standard deviation of the sampling distribution.

Formula: 

\(\displaystyle \bar{x} \pm t\cdot \frac{s}{\sqrt{n}}\)

To find the appropriate t-value for 95% confidence interval: 

\(\displaystyle \alpha =0.05, \frac{0.05}{2}=0.025\)

\(\displaystyle df=n-1= 20-1 = 19\)

Look up \(\displaystyle t_{\frac{\alpha }{2}, 19}\) in t-table and the corresponding t-value = 2.093.

Thus the 95% confidence interval is: 

\(\displaystyle 16.53 \pm 2.093\left(\frac{5.29}{\sqrt{19}}\right)= (13.99 mm Hg, 19.07 mm Hg)\)

For 95% confidence, Z = 1.96.

1) The population M.O.E. =

\(\displaystyle 1.96 \times 7 = 13.720\) 

2) The sample standard deviation = 

\(\displaystyle 7\div\sqrt{36}=1.167\)

The sample M.O.E. = 

\(\displaystyle 1.96\times1.167=2.287\)