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.

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 [tex]71.98 <  \mu <  79.88 [/tex]

The 95% confidence interval of mean pulse rate for adult male is  [tex]62.89 <  \mu <  70.57 [/tex]

The correct option is  C

Step-by-step explanation:

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

       [tex]\= x_1 = \frac{\sum x_i }{n}[/tex]

= >    [tex]\= x_1 = \frac{81 + 74 + \cdots + 59 }{40 }[/tex]

= >    [tex]\= x_1 = 66.73[/tex]

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

      [tex]\sigma_1 = \sqrt{\frac{\sum (x - \= x)^2 }{n-1 } }[/tex]

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

=>   [tex]\sigma_1 = 12.24[/tex]

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

      [tex]\alpha = (100 - 95 ) \%[/tex]

=>   [tex]\alpha = 0.05[/tex]

Generally from the normal distribution table the critical value  of  [tex]\frac{\alpha }{2}[/tex] is  

   [tex]Z_{\frac{\alpha }{2} } =  1.96[/tex]

Generally the margin of error is mathematically represented as  

      [tex]E = Z_{\frac{\alpha }{2} } *  \frac{\sigma_1 }{\sqrt{n} }[/tex]

=>    [tex]E = 1.96 *  \frac{12.4 }{\sqrt{40} }[/tex]

=>    [tex]E =3.84 [/tex]

Generally 95% confidence interval is mathematically represented as  

      [tex]\= x_1 -E <  \mu <  \=x_1  +E[/tex]

=>    [tex]66.73 -3.84 <  \mu < 66.73 +3.84 [/tex]

=>    [tex]62.89 <  \mu <  70.57 [/tex]

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

       [tex]\= x_2 = \frac{\sum x_i }{n}[/tex]

= >    [tex]\= x_2 = \frac{81 + 94 + \cdots + 73 }{40 }[/tex]

= >    [tex]\= x_2 = 75.93[/tex]

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

      [tex]\sigma_2 = \sqrt{\frac{\sum (x - \= x)^2 }{n-1 } }[/tex]

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

=>   [tex]\sigma_2 = 12.73[/tex]

Generally from the normal distribution table the critical value  of  [tex]\frac{\alpha }{2}[/tex] is  

   [tex]Z_{\frac{\alpha }{2} } =  1.96[/tex]

Generally the margin of error is mathematically represented as  

      [tex]E = Z_{\frac{\alpha }{2} } *  \frac{\sigma_2 }{\sqrt{n} }[/tex]

=>    [tex]E = 1.96 *  \frac{12.73 }{\sqrt{40} }[/tex]

=>    [tex]E =3.95 [/tex]

Generally 95% confidence interval is mathematically represented as  

      [tex]\= x_2 -E <  \mu <  \=x_2  +E[/tex]

=>    [tex]75.93 -3.95 <  \mu < 75.93 + 3.95 [/tex]

=>    [tex]71.98 <  \mu <  79.88 [/tex]

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