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marin [14]
3 years ago
12

You claim you are psychic and can predict a coin flip 60% of the time. To test your claim, your friend has you predict a coin fl

ip and then records whether or not you are correct. You do this 200 times and find you are right 135 times. Considering a null hypothesis that you are correctly predict the coin flip 60% of the time, using a hypothesis test, which of the following statements is most accurate?
a. You cannot reject the null at the 10% level of significance.
b. You can reject the null at the 10% level of significance, but cannot reject the null at the 5%, 2%, or 1% level of significance.
c. You can reject the null at the 10% and 5% level of significance, but cannot reject the null at the 2%, or 1% level of significance.
d. You can reject the null at the 10%, 5%, and 2% level of significance, but cannot reject the null at the 1% level of significance.
e. You can reject the null at the 10%, 5%, 2%, and 1% level of significance
Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
5 0

Answer:

You can reject the null at the 10% and 5% level of significance, but cannot reject the null at the 2%, or 1% level of significance. ( D )

Step-by-step explanation:

let : P = 0.6

x = 135 , n = 200

<em>p </em>= 135 / 200 = 0.675

H0 : P = 0.6

Ha : P ≠ 0.6

performing the test statistic = ( 0.675 - 0.6 ) / \sqrt{\frac{0.60(1-0.60)}{200} }

= 0.075 / 0.0346 ≈ 2.17

p-value = 2 * P ( Z > - 2.17 )  ( using excel function ; NORMSDIST )

            = 2 * 0.015  = 0.03  

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Find the dimensions of an open rectangular box with a square base that holds 2000 cubic cm and is constructed with the least bui
Vesnalui [34]
<h3>The dimensions of the given rectangular box are:</h3><h3>L  =   15.874 cm  , B  =  15.874 cm   , H = 7.8937 cm</h3>

Step-by-step explanation:

Let us assume that the dimension of the square base = S x S

Let us assume the height of the rectangular base = H

So, the total area of the open rectangular box  

= Area of the base +  4 x ( Area of the adjacent faces)

=  S x S  +  4 ( S x H)   = S² +  4 SH   ..... (1)

Also, Area of the box  = S x S x H  =  S²H

⇒ S²H = 2000

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Substituting the value of H in (1), we get:

A = S^2 + 4 SH =  S^2 + 4 S(\frac{2000}{S^2}) =  S^2 + (\frac{8000}{S})\\\implies A  =  S^2 + (\frac{8000}{S})

Now, to minimize the area put :

(\frac{dA}{dS} ) = 0 \implies 2S  - \frac{8000}{S^2}  = 0\\\implies S^3 = 4000\\\implies S  = 15.874 \approx 16 cm

Putting the value of S  = 15.874 cm in the value of H , we get:

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