As part of a health study, researchers collected the self-reported weights and t
ID: 3395719 • Letter: A
Question
As part of a health study, researchers collected the self-reported weights and the actual weights from thousands of participants aged 30-45. They take a random sample of responses, which are shown to the right. Construct a 97% confidence interval for the mean difference between self-reported weights and actual weights.
Actual Self-reported 107 112 146 132 120 108 201 195 115 120 220 212 175 170 232 216 114 114 136 115 136 142 198 206 200 192 165 170 134 124 114 107 167 155 131 121 152 146 112 101 148 143 119 125 189 179 220 210 192 186 156 148 167 157 137 130 141 127 221 207 145 132 164 154Explanation / Answer
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=158.5625
Standard deviation( sd1 )=36.741
Sample Size(n1)=32
Mean(x2)=151.75
Standard deviation( sd2 )=35.756
Sample Size(n1)=32
CI = [ ( 158.5625-151.75) ±t a/2 * Sqrt( 1349.901081/32+1278.491536/32)]
= [ (6.81) ± t a/2 * Sqrt( 82.14) ]
= [ (6.81) ± 2.275 * Sqrt( 82.14) ]
= [-13.81 , 27.43]
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