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saw5 [17]
4 years ago
5

The data below were obtained from an experiment were participants were given drinks with or without caffeine and then asked to t

ap their fingers. The data for 20 participants are below. Assume the number of taps per minute is normally distributed. The variance is unknown. Find a 95% CI for μ number of taps. Identify the pivot function used. 246 242 248 245 250 244 252 248 248 247 250 248 246 242 248 244 245 246 250 242
Mathematics
1 answer:
Black_prince [1.1K]4 years ago
8 0

Answer:

The  95% confidence interval is  244.26  <  \mu < 246.95  

The pivot function used is  

           t = \frac{\=x - \mu}{ \frac{\sigma}{\sqrt{n} } }

Step-by-step explanation:

From the question we are told that

  The  data given is 246 242 248 245 250 244 252 248 248 247 250 248 246 242 248 244 245 246 250 242

  The  sample  size is  n=  20

   

Given that the confidence level is 95% then the level of significance is

      \alpha = (100 - 95)\%

      \alpha =  0.05

The  degree of freedom is mathematically represented as

    df =  20 -1

    df =  19

From the student t-distribution table  the critical value of  \frac{\alpha }{2}  is  

          t_{\frac{\alpha }{2} ,  19 } =  2.093

The mean is mathematically represented as

          \= x  = \frac{\sum x_i}{ n}

         \= x  = \frac{246+ 242 +248+245+ 250+ 244+252+ 248 +248 +247+ 250+ 248+ 246+ 242 +248 +244 +245 +246+ 250+ 242}{20}\= x  =  246.6

The standard deviation is mathematically represented as

            \sigma  = \sqrt{\frac{\sum (x_i -  \= x )^2)}{n} }

           \sigma  = \sqrt{\frac{(246- 246.6)^2 +(242- 246.6)^2 +(248- 246.6)^2 + (248- 245)^2+}{20} } \ ..

             \ ...\sqrt{\frac{(250-246.6 )^2+ (244- 246.6)^2+(252- 246.6)^2+ (248- 246.6)^2+ (248- 246.6)^2+}{20} } \ ...

              \ ..\sqrt{\frac{(247- 246.6)^2+ (250- 246.6)^2+ (248-246.6)^2+ (246-246.6)^2+  (242-246.6)^2+  (248-246.6)^2+ (244-246.6)^2+}{20} } \ ...       \sqrt{\frac{ (245-246.6)^2+  (246-246.6)^2+ ( 246-246.6)^2 + ( 250-246.6)^2+ ( 242-246.6)^2 +( 246-246.6)^2+ ( 242-246.6)^2 }{20} }\sigma =  2.87411

The margin of error is mathematically represented as

       E =  t_{\frac{\alpha }{2} , 19} *  \frac{\sigma }{\sqrt{n} }

      E =  2.093 *  \frac{2.87411 }{\sqrt{20} }

      E =  1.345

The 95% confidence interval is mathematically represented as

       \= x  -  E  <  \mu < \= x  +  E

=>   245.6 - 1.345  <  \mu

=>    244.26  <  \mu < 246.95  

The pivot function used is  

           t = \frac{\=x - \mu}{ \frac{\sigma}{\sqrt{n} } }

     

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