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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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Your parallel line is x = - 1. 
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In a study to determine the relationship between self esteem and eating, subjects had been randomly divided into two groups. One
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A sample of 15 commuters in Chicago showed the average of the commuting times was 33.2 minutes. If s = 8.3 minutes, find the 95%
OleMash [197]

Answer:

The 95% confidence interval of the true mean.

(29.4261 ,36.9739)

Step-by-step explanation:

<u>Step :- (i)</u>

Given sample size 'n' =15

sample of the mean x⁻ = 33.2

The standard deviation of the sample 'S' = 8.3

<u>95% of confidence intervals</u>

<u></u>(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )<u></u>

<u>Step:-(ii)</u>

<u>The degrees of freedom γ=n-1 = 15-1=14</u>

The tabulated value t = 1.761 at 0.05 level of significance.

now substitute all possible values, we get

(33.2 - 1.761\frac{8.3}{\sqrt{15} } ,33.2+ 1.761\frac{8}{\sqrt{15} } )

After calculation , we get

(33.2-3.7739 , 33.2+3.7739

(29.4261 ,36.9739)

<u>Conclusion</u>:-

the 95% confidence interval of the true mean.

(29.4261 ,36.9739)

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<span>I think the answer is The function f(x) = 9,000(0.95)x </span>
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