A survey of several 9 to 11 year olds recorded the following amounts spent on a trip to the mall: $23.71,$17.79,$29.87,$18.78,$28.76 Construct the 95% confidence interval for the average amount spent by 9 to 11 year olds on a trip to the mall. Assume the population is approximately normal. Step 3 of 4 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

(16.904 ; 30.660)

Step-by-step explanation:

Given the data :

X = 23.71,17.79,29.87,18.78,28.76

Sample size = 5

Mean = (23.71+17.79+29.87+18.78,+28.76) / 5 = 23.782

Using a calculator, sample standard deviation :

Standard deviation, s = 5.54

Tcritical at α = 0.05, df = 5 - 1 = 4

Using tables ; Tcritical value = 2.7763

Confidence interval :

Mean ± Margin of Error

Margin of Error (MOE) = Tcritical * SE

Margin of Error = Tcritical * s/sqrt (n)

MOE = 2.7763 * 5.54/Sqrt(5) = 6.878

Confidence interval : 23.782 ± 6.878

Lower boundary : 23.782 - 6.878 = 16.904

Upper boundary = 23.782 + 6.878 = 30.66

(16.904 ; 30.660)

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step-by-step explanation:

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step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt