Question

The students answered questions about alcohol use and hangovers, including a count of how many out of a list of 13 possible hangover symptoms they had experienced in the past year. For the 470 men, the mean number of symptoms was 5.4; for the 755 women, it was 5.1. The standard deviation was 3.7 for each of the two samples.

(a) Define the parameter of interest using appropriate notation. The parameter of interest is Correct: Your answer is correct. in mean number of hangover symptoms, out of the 13 listed, experienced by the populations of male and female college students similar to those in the study.

(b) Find the standard error of the difference in means. (Round the answer to four decimal places.) s.e.(x1 - x1) =

(c) Find an approximate 95% confidence interval for the difference in population means. (Round all answers to two decimal places.) to symptoms

(d) On the basis of your interval in part (c), do you think that there is a difference in the population mean number of hangover symptoms experienced by college men and women? There Correct: Your answer is correct. a difference in the mean number of symptoms in the population.

Answer 1

Part a)

Parameter of interest is the difference between mean numbers of hangover symptoms, out of the 13 listed, experienced by the populations of male and female college students similar to those in the study.

Parameter of interest is the difference between two population means (µ₁ - µ₂).

Part b

Standard error for difference is given as below:

SE = S_p²*((1/n1)+(1/n2))

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

We are given n1=470, n2=755, S1=3.7, S2=3.7

S_p² = [(470 – 1)*3.7^2 + (755 – 1)*3.7^2]/(470 + 755 – 2)

S_p² = 13.69

SE = S_p²*((1/n1)+(1/n2))

SE = 13.69*((1/470)+(1/755))

SE = 0.2174

Required standard error = 0.2174

Part c

Confidence interval = (X1bar – X2bar) ± t*SE

We are given

X1bar = 5.4

X2bar = 5.1

(X1bar – X2bar) = 0.3

SE = 0.2174

Confidence level = 95%

DF = n1 + n2 – 2 = 470+755-2 = 1223

Critical t value = 1.9619 (by using t-table)

Confidence interval = 0.3 ± 1.9619*0.2174

Confidence interval = 0.3 ± 0.42651706

Lower limit = 0.3 - 0.42651706 = -0.12651706

Upper limit = 0.3 + 0.42651706 = 0.72651706

Confidence interval = (-0.13, 0.73)

Part d

There is no any significant difference in the population mean number of hangover symptoms experienced by college men and women because the value ‘0’ is lies within above confidence interval.