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skelet666 [1.2K]
3 years ago
5

Two different weight loss programs are being compared to determine their effectiveness. Ten men were assigned to each program (t

hat is, a total of 20 men altogether). Their weight losses (in lbs), after a period of time, are recorded below. We are interested in determining which, if any, of the diets is more effective in terms of average weight loss. Assume weight loss for each diet to be normally distributed.
Diet 1 3.4 10.9 2.8 7.8 0.9 5.2 2.5 10.5 7.1 7.5
Diet 2 11.9 13.1 11.6 6.8 6.8 8.8 12.5 8.6 17.5 10.3
Carry out an appropriate test using a significance level of 0.10.
Mathematics
1 answer:
True [87]3 years ago
5 0

Answer:

WE reject the Null and conclude that one of the drug is more effective than the other.

Step-by-step explanation:

Given :

Diet 1 3.4 10.9 2.8 7.8 0.9 5.2 2.5 10.5 7.1 7.5

Diet 2 11.9 13.1 11.6 6.8 6.8 8.8 12.5 8.6 17.5 10.3

This is a matched pair sample :

The hypothesis :

H0 : μd = 0

H1 : μd ≠ 0

Hence, we intun the difference between the two groups of value :

Difference, d = -8.9,-2.2,-8.8,1,-5.9,-3.6,-10,1.9,-10.4,-2.8

The test statistic :

dbar ÷ (Sd/√n)

dbar = mean of difference = Σd / n = - 49.7 / 10 = - 4.97

Standard deviation of difference, Sd = 4.51

Test statistic :

-4.97 ÷ (4.51/√10)

Test statistic = - 3.485

The sample size, n = 10

df = n - 1 ; 10 - 1 = 9

Critical value (0.10, 9) = 1.833

If Test statistic > |Critical value |

Since 3.486 > 1.833 ; WE reject the Null and conclude that one of the drug is more effective than the other

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∆ ABC is similar to ∆DEF and their areas are respectively 64cm² and 121cm². If EF = 15.4cm then find BC.​
lyudmila [28]

{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}

★ ∆ ABC is similar to ∆DEF

★ Area of triangle ABC = 64cm²

★ Area of triangle DEF = 121cm²

★ Side EF = 15.4 cm

{\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}

★ Side BC

{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}

Since, ∆ ABC is similar to ∆DEF

[ Whenever two traingles are similar, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. ]

\therefore \tt \boxed{  \tt \dfrac{area( \triangle \: ABC )}{area( \triangle \: DEF)} =  { \bigg(\frac{BC}{EF} \bigg)}^{2}   }

❍ <u>Putting the</u><u> values</u>, [Given by the question]

• Area of triangle ABC = 64cm²

• Area of triangle DEF = 121cm²

• Side EF = 15.4 cm

\implies  \tt  \dfrac{64   \: {cm}^{2} }{12 \:  {cm}^{2} }  =  { \bigg( \dfrac{BC}{15.4 \: cm} \bigg) }^{2}

❍ <u>By solving we get,</u>

\implies  \tt    \sqrt{\dfrac{{64 \: cm}^{2} }{ 121 \: {cm}^{2} }}   =   \bigg( \dfrac{BC}{15.4 \: cm} \bigg)

\implies  \tt    \sqrt{\dfrac{{(8 \: cm)}^{2} }{  {(11 \: cm)}^{2} }}   =   \bigg( \dfrac{BC}{15.4 \: cm} \bigg)

\implies  \tt    \dfrac{8 \: cm}{11 \: cm}    =   \dfrac{BC}{15.4 \: cm}

\implies  \tt    \dfrac{8  \: cm \times 15.4 \: cm}{11 \: cm}    =   BC

\implies  \tt    \dfrac{123.2 }{11 } cm   =   BC

\implies  \tt   \purple{  11.2 \:  cm}   =   BC

<u>Hence, BC = 11.2 cm.</u>

{\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}

★ Figure in attachment.

\rule{280pt}{2pt}

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Answer:

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Step-by-step explanation:

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