9. Psychologists at an university compared the levels of alcohol consumption of

Question

9. Psychologists at an university compared the levels of alcohol consumption of male and female freshman students. Each student was asked to estimate the amount of alcohol (beer, wine, or liquor) they consume in a typical week. Summary statistics for 126 males and 184 females are provided in the accompanying table Click here to view the table a. For each gender, find a 95% confidence interval for mean weekly alcohol consumption For males, the 95% confidence interval is ( (Round to the nearest hundredth as needed.) For females, the 95% confidence interval is ( (Round to the nearest hundredth as needed.) b. Prior to sampling, what is the probability that at least one of the two confidence intervals will not contairn the population mean it estimates. Assume that the two intervals are independent. The probability isS (Round to four decimal places as needed.) c. Based on the two confidence intervals, what inference can you make about which gender consumes the most alcohol, on average, per week? 0 A. The females consume the most alcohol, on average, per week O B. The males consume the most alcohol, on average, per week ° C. There is not enough information to answer this question 1: Data Table Sample size, n Mean (ounces), x Standard deviation, s Males 126 16.75 13.58 Females 184 10.77 11.54

Explanation / Answer

PART A.
i.
TRADITIONAL METHOD
given that,
sample mean, x =16.75
standard deviation, s =13.58
sample size, n =126
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 13.58/ sqrt ( 126) )
= 1.21
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 125 d.f is 1.979
margin of error = 1.979 * 1.21
= 2.394
III.
CI = x ± margin of error
confidence interval = [ 16.75 ± 2.394 ]
= [ 14.356 , 19.144 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =16.75
standard deviation, s =13.58
sample size, n =126
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 125 d.f is 1.979
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 16.75 ± t a/2 ( 13.58/ Sqrt ( 126) ]
= [ 16.75-(1.979 * 1.21) , 16.75+(1.979 * 1.21) ]
= [ 14.356 , 19.144 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 14.356 , 19.144 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean

ii.
from standard normal table, two tailed value of |t /2| with n-1 = 183 d.f is 1.973
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 10.77 ± t a/2 ( 11.54/ Sqrt ( 184) ]
= [ 10.77-(1.973 * 0.851) , 10.77+(1.973 * 0.851) ]
= [ 9.091 , 12.449 ]

PART B.
The probability that at least one of the two confidence intervals do not contain the
population mean is,
p = 1 —(0.95x 0.95)
= 1-0.9025
= 0.0975
=
PART C.
the male consumes the most alhocal per week

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