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Elanso [62]
3 years ago
9

Lucy is using a one-sample t ‑test based on a simple random sample of size n = 22 to test the null hypothesis H 0 : μ = 16.000 c

m against the alternative H 1 : μ < 16.000 cm. The sample has mean ¯¯¯ x = 16.218 cm and standard deviation is s = 0.764 cm. Determine the value of the t ‑statistic for this test. Give your answer to three decimal places.
Mathematics
1 answer:
slega [8]3 years ago
4 0

Answer:

The value of test statistic is 1.338

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 16.000

Sample mean, \bar{x} = 16.218

Sample size, n = 22

Alpha, α = 0.05

Sample standard deviation, s = 0.764

First, we design the null and the alternate hypothesis

H_{0}: \mu = 16.000\text{ cm}\\H_A: \mu < 16.000\text{ cm}

We use one-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{16.218 - 16.000}{\frac{0.764}{\sqrt{22}} } = 1.338

Thus, the value of test statistic is 1.338

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The volume of the sawdust pile is 11,039.06 ft³.

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The shape of the sawdust pile would be a cone. The diameter is always proportional to its height.

Diameter = k x height

where k is equivalent to the rate of increase of the diameter with respect to the increase in height.

25= k20

k = 25 / 20

k = 1.25

Diameter when height is 30 ft = 1.25x 30 = 35.50 feet

Volume of the sawdust = \frac{1}{3}πr²h

Where:

  • π = 3.14
  • r = radius = diameter / 2 = 35.50 / 2 = 18.75
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\frac{1}{3} x 3.14 x 18.75² x 30 = 11,039.06 ft³

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the answer is below

Step-by-step explanation:

We have the total number of possible samples = 4 * 4 = 16 (As 4 choices for each value)

in addition we have to as all sample occur with equal probability, probability of each sample = 1/16, below is samling distribution of mean

x1 x2 probabilityP(x1,x2) sample mean

190 190    1/16                 190

190 100    1/16                 145

190 272    1/16                 231

190 74    1/16                 132

100 190    1/16                 145

100 100    1/16                 100

100 272    1/16                 186

100 74    1/16                        87

272 190    1/16                 231

272 100    1/16                 186

272 272    1/16                 272

272 74    1/16                 173

74 190    1/16                 132

74 100    1/16                 87

74 272    1/16                 173

74 74    1/16                 74

Summarizing above with adding duplicate values

sample mean   probability

       74              1/16

      87               1/8

    100                1/16

    132                1/8

    145                1/8

    173                1/8

   186                1/8

   190               1/16

    231               1/8

   272               1/16

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