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I need the solution of one question 1) Supposed you are interested in determinin

ID: 2926679 • Letter: I

Question

I need the solution of one question

1) Supposed you are interested in determining if the mean number of stolen items sold differs between offenders who used social media networks and those who used pawnshops. You interviewed 75 offenders who used social media networks to get rid of stolen goods and found that the mean number of stolen items sold in a month was 2.5 (standard deviation is 0.45). Construct an 85 percent confidence interval AND a 99 percent confidence interval around this point estimate (10 points).

Explanation / Answer

(a)

n = 75     

x-bar = 2.5     

s = 0.45     

% = 85     

Standard Error, SE = s/n =    0.45/75 = 0.051961524

Degrees of freedom = n - 1 =   75 -1 = 74   

t- score = 1.454631687     

Width of the confidence interval = t * SE =     1.45463168727951 * 0.0519615242270663 = 0.07558488

Lower Limit of the confidence interval = x-bar - width =      2.5 - 0.0755848796600328 = 2.42441512

Upper Limit of the confidence interval = x-bar + width =      2.5 + 0.0755848796600328 = 2.57558488

(b)

n = 75     

x-bar = 2.5     

s = 0.45     

% = 99     

Standard Error, SE = s/n =    0.45/75 = 0.051961524

Degrees of freedom = n - 1 =   75 -1 = 74   

t- score = 2.643912849     

Width of the confidence interval = t * SE =     2.64391284872332 * 0.0519615242270663 = 0.137381742

Lower Limit of the confidence interval = x-bar - width =      2.5 - 0.137381741543189 = 2.362618258

Upper Limit of the confidence interval = x-bar + width =      2.5 + 0.137381741543189 = 2.637381742

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