Question

Refer to the accompanying data set and construct a ​% confidence interval estimate of the mean pulse rate of adult​ females; then do the same for adult males. Compare the results. Click the icon to view the pulse rates for adult females and adult males. Construct a ​% confidence interval of the mean pulse rate for adult females. nothing bpm nothing bpm ​(Round to one decimal place as​ needed.) Construct a ​% confidence interval of the mean pulse rate for adult males. nothing bpm nothing bpm ​(Round to one decimal place as​ needed.) Compare the results.
A. The confidence intervals​ overlap, so it appears that there is no significant difference in mean pulse rates between adult females and adult males.
B. The confidence intervals​ overlap, so it appears that adult males have a significantly higher mean pulse rate than adult females.
C. The confidence intervals do not​ overlap, so it appears that adult females have a significantly higher mean pulse rate than adult males.
D. The confidence intervals do not​ overlap, so it appears that there is no significant difference in mean pulse rates between adult females and adult males.

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

The 95% confidence interval of mean pulse rate for adult female is $71.98 < \mu < 79.88$

The 95% confidence interval of mean pulse rate for adult male is  $62.89 < \mu < 70.57$

The correct option is  C

Step-by-step explanation:

  Generally the sample mean for Male pulse rate is mathematically represented as

       $\= x_1 = \frac{\sum x_i }{n}$

= >    $\= x_1 = \frac{81 + 74 + \cdots + 59 }{40 }$

= >    $\= x_1 = 66.73$

Generally the standard deviation for male pulse rate is mathematically represented as

      $\sigma_1 = \sqrt{\frac{\sum (x - \= x)^2 }{n-1 } }$

=>   $\sigma_1 = \sqrt{\frac{\sum (81 - 66.73 )^2 + (74 - 66.73 )^2+ \cdot + (59 - 66.73 )^2 }{40-1 } }$

=>   $\sigma_1 = 12.24$

From the question we are told the confidence level is  95% , hence the level of significance is    

      $\alpha = (100 - 95 ) \%$

=>   $\alpha = 0.05$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as  

      $E = Z_{\frac{\alpha }{2} } * \frac{\sigma_1 }{\sqrt{n} }$

=>    $E = 1.96 * \frac{12.4 }{\sqrt{40} }$

=>    $E =3.84$

Generally 95% confidence interval is mathematically represented as  

      $\= x_1 -E < \mu < \=x_1 +E$

=>    $66.73 -3.84 < \mu < 66.73 +3.84$

=>    $62.89 < \mu < 70.57$

  Generally the sample mean for Female pulse rate is mathematically represented as

       $\= x_2 = \frac{\sum x_i }{n}$

= >    $\= x_2 = \frac{81 + 94 + \cdots + 73 }{40 }$

= >    $\= x_2 = 75.93$

Generally the standard deviation for Female  pulse rate is mathematically represented as

      $\sigma_2 = \sqrt{\frac{\sum (x - \= x)^2 }{n-1 } }$

=>   $\sigma_2 = \sqrt{\frac{\sum (81 - 66.73 )^2 + (94 - 66.73 )^2+ \cdot + (73 - 66.73 )^2 }{40 } }$

=>   $\sigma_2 = 12.73$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as  

      $E = Z_{\frac{\alpha }{2} } * \frac{\sigma_2 }{\sqrt{n} }$

=>    $E = 1.96 * \frac{12.73 }{\sqrt{40} }$

=>    $E =3.95$

Generally 95% confidence interval is mathematically represented as  

      $\= x_2 -E < \mu < \=x_2 +E$

=>    $75.93 -3.95 < \mu < 75.93 + 3.95$

=>    $71.98 < \mu < 79.88$