Question

Two random samples of 30 individuals were selected. One sample participated in an activity which simulates hard work. The average breath rate of these individuals was 17 breaths per minute. The other sample did some normal walking. The mean breath rate of these individuals was 14. Find the 90% confidence interval of the difference in the breath rates if the population standard deviation was 4.2 for breath rate per minute.

Answer 1

Answer:

$(17-14) -1.64 \sqrt{\frac{4.2^2}{30} +\frac{4.2^2}{30}}=1.222$

$(17-14) +1.64 \sqrt{\frac{4.2^2}{30} +\frac{4.2^2}{30}}=4.778$

So then we can conclude at 90% of confidence that the difference in the two means is between 1.222 and 4.778 beats per minute

Step-by-step explanation:

For this case we have the following info given:

$\bar X_1 = 17$ sample mean for the first sample

$\bar X_2 = 14$ sample mean for the second sample

$\sigma =4.2$ represent the population deviation

$n_1= n_2 =30$ represent the sample size ofr each case

We can construct the confidence interval for the difference of means with the following formula:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{\sigma^2_1}{n_1} +\frac{\sigma^2_2}{n_2}}$

And the confidence for this case is 90% or 0.9 so then the significance level is $\alpha=1-0.9=0.1$ and $\alpha/2 =0.05$ the critical value for this case is:

$z_{\alpha/2}=1.64$

And replacing we got:

$(17-14) -1.64 \sqrt{\frac{4.2^2}{30} +\frac{4.2^2}{30}}=1.222$

$(17-14) +1.64 \sqrt{\frac{4.2^2}{30} +\frac{4.2^2}{30}}=4.778$

So then we can conclude at 90% of confidence that the difference in the two means is between 1.222 and 4.778 beats per minute