Question

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

Answer 1

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

The pivot function used is

`t = (\=x - \mu)/( (\sigma)/(√(n) ) )`

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
`(\alpha )/(2)` is

`t_{(\alpha )/(2) , 19 } = 2.093`

The mean is mathematically represented as

`\= x = (\sum x_i)/( n)`

`\= x = (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{(\sum (x_i - \= x )^2))/(n) }`

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

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

`\ ..\sqrt{((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{( (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_{(\alpha )/(2) , 19} * (\sigma )/(√(n) )`

`E = 2.093 * (2.87411 )/(√(20) )`

`E = 1.345`

The 95% confidence interval is mathematically represented as

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

=>
`245.6 - 1.345 < \mu <245.6 + 1.345`

=>
`244.26 < \mu < 246.95`

The pivot function used is

`t = (\=x - \mu)/( (\sigma)/(√(n) ) )`