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skelet666 [1.2K]
2 years ago
14

You know that y is a normally distributed variable with a variance of 9. You do not know its mean. You collect some data. For ea

ch sample below, form the 95% confidence interval and test the null hypothesis of the mean equaling 2.
a. (8,1,5)
b. (8,1,5,-4,-8)
c. (8,1,5,-4,-8,4,8,5)
Mathematics
1 answer:
Dima020 [189]2 years ago
8 0

Answer:

a)

CI = ( 1.2718, 8.0614 )

Since value 2 is contained within Confidence Interval, fail to Reject H₀

b)

CI = ( -2.2296, 3.0296 )

Since value 2 is contained within Confidence Interval, fail to Reject H₀

c)

CI = ( 0.2962, 4.4538 )

Since value 2 is contained within Confidence Interval, fail to Reject H₀

Step-by-step explanation:

Given that;

Variance σ² = 9

so standard deviation σ = √9 = 3

with 95% confidence interval in each case. thus, The Z value for 95% confidence is Z = 1.96

Null Hypotheses  H₀ : μ = 2

Now

a)  (8,1,5)

sample mean x" = (8+1+5)  / 3 = 14/3 = 4.6666

n = 3

Confidence Interval CI = x" ± ( z × σ/√n )

CI = 4.6666 ± ( 1.96 × 3/√3 )

CI = 4.6666 ± 3.3948

so CI = ( 1.2718, 8.0614 )

Since value 2 is contained within Confidence Interval, fail to Reject H₀

b)  (8,1,5,-4,-8)  

sample mean X" =  (8 +1 + 5 - 4 - 8)  / 5 = 2 / 5 = 0.4

n = 5  

Confidence Interval CI = x" ± ( z × σ/√n )

CI = 0.4 ± ( 1.96 × 3/√5 )

CI = 0.4 ± 2.6296

so CI = ( -2.2296, 3.0296 )

Since value 2 is contained within Confidence Interval, fail to Reject H₀

c)  (8,1,5,-4,-8,4,8,5)

sample mean x" = (8 + 1 + 5 - 4 - 8 + 4 + 8 + 5)   / 8 = 19 / 8 = 2.375

n = 8

Confidence Interval CI = x" ± ( z × σ/√n )

CI = 2.375 ± ( 1.96 × 3/√8 )

CI = 2.375 ± 2.0788

so CI = ( 0.2962, 4.4538 )

Since value 2 is contained within Confidence Interval, fail to Reject H₀

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Andreas93 [3]

Answer:

10 boxes in the top row.

192 boxes in entire display.

Step-by-step explanation:

Let n be the number of rows.

Given:

Total number of rows = 8

Boxes in bottom row = 38

And each row has four fewer boxes than the row below it.

Solution:

Part A:

We know that the bottom row has 38 boxes and each row has four fewer boxes than the row below it.

Using below function for determining the number of boxes in each rows.

f(n)=38-(8-n)4

Where:

n = Number of row.

We need to find the boxes in the first row.

So, substitute n = 1 in above function.

f(1)=38-(8-1)4

f(1)=38-7\times 4

f(1)=38-28

f(1)=10

Therefore, 10 boxes in the top row.

Part B:

Total boxes in the entire display.

Using formula as given below to determine the sum of the boxes in entire display.

S_{n} = \frac{n}{2}(f(1)+f(n))

Substitute n = 8, f(1) = 10 and f(8) = 38 in the above equation.

S_{8} = \frac{8}{2}(10+38)

S_{8} = 4(48)

S_{8} = 4\times 48

S_{8} = 192\ boxes

Therefore, 192 boxes in the entire display.

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