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Dennis_Churaev [7]

2 years ago

13

Assume that you want to test the claim that the paired sample data come from a population for which the mean difference is μd =

0. Compute the value of the t test statistic. Round intermediate calculations to four decimal places as needed and final answers to three decimal places as needed.
Subject A B C D E F G H I
Before 168 180 157 132 202 124 190 210 171
After 162 178 145 125 171 126 180 195 163

1 answer:

svlad2 [7]2 years ago

8 0

Answer:

t = 3.156

Step-by-step explanation:

N = 9

Please check attachment for the table

ED = 89

ED² = 1587

Bar D = 89/9

= 9.8889

The standard deviation s

= √14283-7921/81

= √78.5432

S = 8.8625

The test statistic

= 9.8889/8.8625/√9-1

= 9.8889/3.1334

= 3.156

From the calculations above, the calculated test statistic t is = 3.156.

Please check the attachment for a more detailed solution to this problem.

Thank you!

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Answer:

see explanation

Step-by-step explanation:

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bixtya [17]

Answer: $\bold{(1)\ \dfrac{19,683}{64}\qquad (2)\ 16}$

<u>Step-by-step explanation:</u>

(1) (12, 18, 27, ...)

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$a_n=a_o(r)^{n-1}\\\\Given:a_o=12,\ r=\dfrac{3}{2}\\\\\\Equation:\\a_n =12\bigg(\dfrac{3}{2}\bigg)^{n-1}\\\\\\\\9th\ term:\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{9-1}\\\\\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{8}\\\\\\.\quad =\large\boxed{\dfrac{19643}{64}}$

$(2)\qquad \bigg(\dfrac{1}{16},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{2}\bigg)\\\\\\\text{The common ratio is}:\\\\r=\dfrac{a_{n+1}}{a_n}\quad r=\dfrac{\frac{1}{8}}{\frac{1}{16}}=\boxed{2}\quad \rightarrow \quad r=\dfrac{\frac{1}{4}}{\frac{1}{8}}=\boxed{2}$

The equation is:

$a_n=a_o(r)^{n-1}\\\\Given:a_o=\dfrac{1}{16},\ r=2\\\\\\Equation:\\a_n =\dfrac{1}{16}(2)^{n-1}\\\\\\\\9th\ term:\\a_9=\dfrac{1}{16}(2)^{9-1}\\\\\\a_9=\dfrac{1}{16}(2)^{8}\\\\\\.\quad =\large\boxed{16}$

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