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tangare [24]
3 years ago
10

A chi-square test for homogeneity was conducted to investigate whether the four high schools in a school district have different

absentee rates for each of four grade levels. The chi-square test statistic and p -value of the test were 19.02 and 0.025, respectively. Which of the following is the correct interpretation of the p -value in the context of the test?
Mathematics
2 answers:
pentagon [3]3 years ago
8 0

Answer:

Assuming that each high school has the same absentee rate for each grade level, there is a 2.5 percent chance of finding a test statistic 19.02 or larger

Step-by-step explanation:

Just took the test and it was the right answer

bogdanovich [222]3 years ago
3 0

Answer:

Option B. Assuming that each high school has the same absentee rate for each grade level, there is a 2.5 percent chance of finding a test statistic 19.02 or larger

Step-by-step explanation:

Since p value is probability of getting the sample statistic as or more extreme given null hypothesis is true

(feel free to leave a thanks and/or 5 star rating)

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Delvig [45]

Answer:

x/i=22

Step-by-step explanation:

x/33=2/3*i

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x=2/3*33*i

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x/i=2/3*33

x/i=22

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2 years ago
Solve this problem thanks
vovikov84 [41]

The three missing lengths are the left hypotenuse, x, the middle altitude, y, and the right hypotenuse, z.


9/y = y/16


y^2 = 9 * 16


y^2 = 144


y = 12


9^2 + 12^2 = x^2


x^2 = 225


x = 15


12^2 + 16^2 = z^2


z^2 = 400


z = 20


From left to right, the sides measure 15, 12, and 20 units.

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3 years ago
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Verify please dont understand
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3 years ago
Trent began his math work at 5:29 p.m. He finished at 6:04 p.m. How many minutes did Trent spend on his math homework?
Morgarella [4.7K]

Answer: I am pretty sure it is 35 min

Step-by-step explanation:

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3 years ago
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What is the variance of a portfolio invested 21 percent each in a and b and 58 percent in c?
victus00 [196]
Given the following information:

\begin{tabular}&#10;{|p{1.5cm}|p{1.5cm}|p{1.2cm}|p{1.2cm}|p{1.2cm}|}&#10;\multicolumn{1}{|p{1.5cm}|}{State of economy}\multicolumn{1}{|p{2.6cm}|}{Probability of State of economy}\multicolumn{3}{|p{4.8cm}|}{Rate of Return if State Occurs}\\[1ex] &#10;\multicolumn{1}{|p{1.5cm}|}{}\multicolumn{1}{|p{2.6cm}|}{}\multicolumn{1}{|c|}{Stock A}&StockB&Stock C\\[2ex]&#10;\multicolumn{1}{|p{1.5cm}|}{Boom}\multicolumn{1}{|p{2.6cm}|}{0.66}\multicolumn{1}{|p{1.27cm}|}{0.09}&0.03&0.34\\&#10;\end{tabular}
\begin{tabular}&#10;{|p{1.5cm}|p{1.5cm}|p{1.2cm}|p{1.2cm}|p{1.2cm}|}&#10;\multicolumn{1}{|p{1.5cm}|}{Bust}\multicolumn{1}{|p{2.6cm}|}{0.34}\multicolumn{1}{|p{1.27cm}|}{0.23}&0.29&-0.14\\&#10;\end{tabular}

Part A:

The expected return on an equally weighted portfolio of these three stocks is given by:

0.66[0.33 (0.09) + 0.33 (0.03) + 0.33(0.34)] \\ +0.34[0.33 (0.23) + 0.33(0.29) +0.33(-0.14)] \\  \\ =0.66(0.0297 + 0.0099 + 0.1122)+0.34(0.0759+0.0957-0.0462) \\  \\ =0.66(0.1518)+0.34(0.1254)=0.1002+0.0426=0.1428=\bold{14.28\%}



Part B:

Value of a portfolio invested 21 percent each in A and B and 58 percent in C is given by

For boom: 0.21(0.09) + 0.21(0.03) + 0.58(0.34) = 0.0189 + 0.0063 + 0.1972 = 0.2224 or 22.24%.

For bust: = 0.21(0.23) + 0.21(0.29) + 0.58(-0.14) = 0.0483 + 0.0609 - 0.0812 = 0.028 or 2.8%

Expected return = 0.66(0.2224) + 0.34(0.028) = 0.1468 + 0.00952 = 0.1563 or 15.63%

The variance is given by

0.66(0.2224-0.1563)^2+0.34(0.028-0.1563)^2 \\  \\ =0.66(0.0661)^2+0.34(-0.1283)^2=0.66(0.00437)+0.34(0.01646) \\  \\ =0.00288+0.0056=\bold{0.00848}
4 0
3 years ago
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