cardiac patients

[logistic_guy]

Many cardiac patients wear an implanted pacemaker to control their heartbeat. A plastic connector module mounts on the top of the pacemaker. Assuming a standard deviation of \(\displaystyle 0.0015\) inch and an approximately normal distribution, find a \(\displaystyle 95\%\) confidence interval for the mean of the depths of all connector modules made by a certain manufacturing company. A random sample of \(\displaystyle 75\) modules has an average depth of \(\displaystyle 0.310\) inch.

 

[khansaheb]

Many cardiac patients wear an implanted pacemaker to control their heartbeat. A plastic connector module mounts on the top of the pacemaker. Assuming a standard deviation of \(\displaystyle 0.0015\) inch and an approximately normal distribution, find a \(\displaystyle 95\%\) confidence interval for the mean of the depths of all connector modules made by a certain manufacturing company. A random sample of \(\displaystyle 75\) modules has an average depth of \(\displaystyle 0.310\) inch.

show us your effort/s to solve this problem.

 

[logistic_guy]

Theorem.

The confidence interval is given by:

\(\displaystyle \left(\bar{x} - z_{\alpha/2}\frac{\sigma}{\sqrt{n}}, \ \bar{x} + z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\right)\)

where \(\displaystyle z_{\alpha/2}\) is the \(\displaystyle z\)-value leaving an area of \(\displaystyle \alpha/2\) to the right.

We are given:

\(\displaystyle 1 - \alpha = 0.95\)
\(\displaystyle \alpha = 1 - 0.95 = 0.05\)

Then,

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

\(\displaystyle z_{0.025}\) (leaving an area of \(\displaystyle 0.025\) to the right) = \(\displaystyle z_{0.975}\).

Looking at the normal table, we find that \(\displaystyle z_{0.975} = 1.96\).

Then, the \(\displaystyle 95\%\) confidence interval is:

\(\displaystyle \left(\bar{x} - z_{\alpha/2}\frac{\sigma}{\sqrt{n}}, \ \bar{x} + z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\right)\)

\(\displaystyle \left(0.310 - 1.96\frac{0.0015}{\sqrt{75}}, \ 0.310 + 1.96\frac{0.0015}{\sqrt{75}}\right)\)

\(\displaystyle \left(0.3097, \ 0.3103\right)\)

 

[khansaheb]

find a 95%\displaystyle 95\%95% confidence interval for the mean of the depths of all connector modules made by a certain manufacturing company

Please be specific with your answer to get full credit.....

 

[logistic_guy]

Please be specific with your answer to get full credit.....

Do you mean that it is better to write the interval as:

\(\displaystyle 0.3097 < \mu < 0.3103\)

?

 

[khansaheb]

Do you mean that it is better to write the interval as:

\(\displaystyle 0.3097 < \mu < 0.3103\)

?

Yes and include units.....