Question

Healthy women aged 18 to 40 participated in a study of eating habits. Subjects were given bags of potato chips and bottled water and invited to snack freely. Interviews showed that some women were trying to restrain their diet out of concern about their weight. How much effect did these good intentions have on their eating habits? Here are the data on grams of potato chips consumed (note that the study report gave the standard error of the mean rather than the standard deviation): Group Unrestrained Restrained SEM 62.83 8.27 31 The standard error of the difference of sample means (+0.0001) is: The critical value ( 0.001) from the distribution for confidence interval 95 % using the conservative degrees of freedom is: Give a 95 % confidence interval ( 0.01) that describes the effect of restraint: to Based on this interval, is there a significant difference between the two groups eBook No. Yes

Answer 1

std error of mean = s/√n

s=SEM*√n

Level of Significance ,    α =    0.05
      
mean of sample 1,    x̅1=   62.83
standard deviation of sample 1,   s1 =8.27*√9 = 24.81
size of sample 1,    n1=   9
      
mean of sample 2,    x̅2=   31
standard deviation of sample 2,   s2 = 8*√ 11 = 26.533
size of sample 2,    n2=   11
      
difference in sample means =    x̅1-x̅2 =    31.83
      
std error of difference, SE =    √(s1²/n1+s2²/n2) =    11.5879

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DF = min(n1-1 , n2-1 )=   8
α=0.05

t-critical value , t* =    2.306 [(excel formula =t.inv(α/2,df)]

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std error , SE =     11.5879      
margin of error, E =    t*SE =    26.722

difference of means =    x̅1-x̅2 =    31.83
confidence interval is               
Interval Lower Limit=   (x̅1-x̅2) - E =    5.11
Interval Upper Limit=   (x̅1-x̅2) + E =    58.55