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adelina 88 [10]
2 years ago
5

Suppose we have two thermometers. One thermometer is very precise but is delicate and heavy (X). We have another thermometer tha

t is much cheaper and lighter, but of unknown precision (Y). We would like to know if we can (reliably) bring the lighter thermometer with us into the field. So, we set up an experiment where we expose both thermometers to 31 different temperatures and measure the temperature with each.
We get the following observations:
x = 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120
y = 0.02, 3.99, 7.91, 12.03, 16.09, 20.00, 23.98, 28.09, 31.94, 36.03, 40.00, 44.05, 47.95, 52.00, 55.87, 59.90, 63.91, 67.95, 72.11, 76.02, 80.01, 84.10, 88.06, 91.74, 96.02, 99.95, 103.87, 108.01, 111.99, 116.04, 120.03
We want to test if these thermometers seem to be measuring the same temperatures. Let's use the threshold α=0.1.
Answer all questions up to 3 decimals
(a) Write down the appropriate hypothesis tests for β1.
H0: β1(≠ / = / > / <) 1 and Ha: β1( ≤ / = / > / ≠ / ≥ / <) 1.
(b) The test statistic is _____. (Use 2 decimal places)
(c) The p-value is _____. (Use 4 decimal places)

Mathematics
1 answer:
makvit [3.9K]2 years ago
8 0

Answer:

Step-by-step explanation:

(a)   H0: μ_D=0

      Ha: μ_D ≠ 0

b) Find attached the solution

(c)  By technology,

p - value = 0.4437

Hence, the p-value is 0.4437

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NOTE THIS IS AN EXAMPLE:

Let t = time, s = ostrich, and g = giraffe.

Here's what we know:

s = g + 5 (an ostrich is 5 mph faster than a giraffe)

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We have a value for s, so plug it into the first equation:

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Isolate g so that we can plug that variable value into the equation:

g = 6/t

so that gives us:

(6/t + 5)t = 7

Distribute:

6 + 5t = 7

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Now that we have a value for time, we can plug them into our equations:

1/5 g = 6

multiply by 5:

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