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Bezzdna [24]
3 years ago
8

Twenty students in Class A and 20 students in Class B were asked how many hours they took to prepare for an exam. The data sets

represent their answers. Class A: {2, 5, 7, 6, 4, 3, 8, 7, 4, 5, 7, 6, 3, 5, 4, 2, 4, 6, 3, 5} Class B: {3, 7, 6, 4, 3, 2, 4, 5, 6, 7, 2, 2, 2, 3, 4, 5, 2, 2, 5, 6} Which statement is true for the data sets? The mean study time of students in Class A is less than students in Class B. The mean study time of students in Class B is less than students in Class A. The median study time of students in Class B is greater than students in Class A. The range of study time of students in Class A is less than students in Class B. The mean and median study time of students in Class A and Class B is equal.
Mathematics
1 answer:
serg [7]3 years ago
7 0
CLASS A
Mean: 4.8
Median: 5
Range: 6

CLASS B
Mean: 4
Median: 4
Range: 5

Therefore, the mean study time of students in Class B is less than that of students in Class A.
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-4.2

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0.5 - 4.7= -4.2

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Solnce55 [7]

Answer:

A = 8

B = 6

C = 4

D = 7

Top = 29

Side = 20

Step-by-step explanation:

To get A on the bottom side row you divide 32 by 4 and get 8.

To get D, on the first side row, subtract A, 8, now it's 21, 21 divided by 3 is 7.

To get B, on the first top row, subtract A and D, now divide by 2 and you get 6.

To get C, on the third side row, subtract B and divide by 2 and you get 4.

4 0
3 years ago
A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro
-Dominant- [34]

Answer:

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

Step-by-step explanation:

Data given and notation  

\bar X=1.6 represent the sample mean

s=0.46 represent the sample deviation

n=32 sample size  

\mu_o =1.35 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:  

Null hypothesis:\mu =1.35  

Alternative hypothesis:\mu \neq 1.35  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

P-value  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

5 0
3 years ago
Alexis G. has a cell phone plan that charges $0.05 per minute plus a monthly fee of $25.00. She budgets $35.50 per month for tot
jeyben [28]
The Maximum amount of minutes she could use would be 210

Formula:
0.05x + 25 <= 35.50
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