Eleven males and 11 females were randomly sampled and administered a projective

Question

Eleven males and 11 females were randomly sampled and administered a projective technique to assess their willingness to self-disclose information about themselves. Assume willingness to self-disclose scores are normally distributed. What is the best point estimate of the difference between males and females in willingness to self-disclose information about themselves? Construct the 90% confidence interval for the mean difference between males and females in willingness to self-disclose information about themselves? Briefly interpret both the point estimate and the confidence interval. The scores by gender were as follows.

Females42 38 46 37 42 47 36 39 45 47 38

Males 30 40 38 37 37 35 32 34 45 33 39 40

Explanation / Answer

a.

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)=41.5455
Standard deviation( sd1 )=4.18
Sample Size(n1)=11
Mean(x2)=36.6667
Standard deviation( sd2 )=4.1414
Sample Size(n2)=12
CI = [ ( 41.5455-36.6667) ±t a/2 * Sqrt( 17.4724/11+17.15119396/12)]
= [ (4.8788) ± t a/2 * Sqrt( 3.0177) ]
= [ (4.8788) ± 1.812 * Sqrt( 3.0177) ]
= [1.7311 , 8.0265]

Interpretations:
1) We are 90% sure that the interval [1.7311 , 8.0265] contains the difference between
true population mean U1 - U2
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the difference between
true population mean U1 - U2
3) Since this Cl does contain a zero we can conclude at 0.10 true mean
difference is zero  

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