A study on students drinking habits asks a random sample of 124"non-greek" UF students how many alcoholic beverages they haveconsumed in the past week. The sample reveals an average of 3.66alcoholic drinks, with a standard deviation of 2.82. Construct a90% confidence interval for the true average number of alcoholicdrinks all UF "non-greek" students have in a one week period.

90% confidence interval for the true average number of alcoholic drinks all UF "non-greek" students have in a one week period is between a lower limit of 3.24 and an upper limit of 4.08.

Explanation:

Confidence interval is given as mean +/- margin of error (E)

mean = 3.66

sd = 2.82

n = 124

degree of freedom = n-1 = 124-1 = 123

confidence interval = 90%

Significance level = 100 - 90 = 10%

Critical value (t) corresponding to 123 degrees of freedom and 10% significance level is 1.6577

E = t×sd/√n = 1.6577×2.82/√124 = 0.42

Lower limit = mean - E = 3.66 - 0.42 = 3.24

Upper limit = mean + E = 3.66 + 0.42 = 4.08

90% confidence interval is (3.24, 4.08)

andromache andromache · Answer 2 · 2022-03-03T16:53:08+01:00

The 90% confidence interval for the true average number of alcoholic drinks all UF "non-greek" students have in a one week period should be

the lower limit of 3.24 and an upper limit of 4.08.

Calculation of the confidence interval:

Since

= mean +/- margin of error (E)

Here

mean = 3.66

sd = 2.82

n = 124

Now

degree of freedom = n-1

= 124-1

= 123

Now

confidence interval = 90%

So,

Significance level = 100 - 90

= 10%

Now

E = t×sd/√n

= 1.6577×2.82/√124

= 0.42

So,

Lower limit = mean - E = 3.66 - 0.42 = 3.24

Upper limit = mean + E = 3.66 + 0.42 = 4.08

hence, The 90% confidence interval for the true average number of alcoholic drinks all UF "non-greek" students have in a one week period should be the lower limit of 3.24 and an upper limit of 4.08.

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