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KIM [24]
3 years ago
7

The sample consisted of 50 night students, with a sample mean GPA of 3.02 and a standard deviation of 0.08, and 25 day students,

with a sample mean GPA of 3.04 and a standard deviation of 0.06. The test statistic is:
Mathematics
1 answer:
bulgar [2K]3 years ago
7 0

Answer: The test statistic is t= -0.90.

Step-by-step explanation:

Since we have given that

n₁ = 50

n₂ = 25

\bar{x_1}=3.02\\\\\bar{x_2}=3.04\\\\\sigma_1=0.08\\\\\sigma_2=0.06

So, s would be

s=\sqrt{\dfrac{n_1\sigma_1^2+n_2\sigma_2^2}{n_1+n_2-2}}\\\\s=\sqrt{\dfrac{50\times 0.08^2+25\times 0.06^2}{50+25-2}}\\\\s=0.075

So, the value of test statistic would be

t=\dfrac{\bar{x_1}-\bar{x_2}}{s(\dfrac{1}{n_1}+\dfrac{1}{n_2})}\\t=\dfrac{3.02-3.04}{0.074(\dfrac{1}{50}+\dfrac{1}{25})}\\\\t=\dfrac{-0.04}{0.074(0.02+0.04)}\\\\t=\dfrac{-0.04}{0.044}\\\\t=-0.90

Hence, the test statistic is t= -0.90.

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Answer:

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8 0
3 years ago
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3 years ago
The U.S. Federal Seed Act establishes germination rates for various fruit and vegetable seeds. Watermelon seeds are to meet a 70
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Answer:

Following are the responses to the given question:

Step-by-step explanation:

Level of Significance,\alpha =    0.05

Sample Size, n =    120          

Sample Proportion,   \hat{P} = 0.600      

z -value =   Z_{\frac{\alpha}{2}} =  1.960  \ \  [excel formula =NORMSINV(\frac{\alpha}{2})]

Standard ErrorSE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} =    0.0447        

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3 years ago
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lys-0071 [83]

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3 0
3 years ago
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