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uysha [10]
2 years ago
11

A survey asked 20 mobile phone dealers how many mobile phones were sold during the previous week. Their answers are displayed in

the Box-and-Whisker Plot and the Stem-and-Leaf Plot shown. Which graph can be used to find the mode? What is the mode of the data set? A. Stem-and-Leaf Plot, Mode: 34 B. Stem-and-Leaf Plot, Mode: 42 C. Box-and-Whisker Plot, Mode: 43 D. Box-and-Whisker Plot, Mode: 38
ASAP 100 POINT AND BRAINLIST
Mathematics
1 answer:
Sidana [21]2 years ago
8 0

The best type of chart for quickly deriving the mode of a sample data is called the Stem-and-Leaf Plot.

<h3>What is a Stem-and-Leap Plot?</h3>

In statistics, the Stem-and-Leaf Plot is an easy-to-make easy-to-read kind of graph that is derived from the table holding the sample data.

The Box-and-Whisker Plot on the other hand is best for visually depicting the five-number summary of any set of data, which are:

  • Minimum
  • First Quartile
  • Median (Second Quartile)
  • Third Quartile; and
  • Maximum.

It is to be noted that referenced plots are not indicated hence, the general answer.

Learn more about Stem-and-Leaf Plot at:
brainly.com/question/8649311
#SPJ1

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In 2008, the average household debt service ratio for homeowners was 13.2. The household debt service ratio is the ratio of debt
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Answer:

t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436    

df=n-1=44-1=43  

p_v =P(t_{(43)}>1.436)=0.079  

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

Step-by-step explanation:

Information given

\bar X=13.88 represent the sample mean

s=3.14 represent the sample standard deviation

n=44 sample size  

\mu_o =13.2 represent the value that we want to test

\alpha=0.05 represent the significance level

t would represent the statistic (variable of interest)  

p_v represent the p value for the test

Hypothesis to test

We want to conduct a hypothesis in order to check if the true mean has increased from 2008 , and the system of hypothesi are:  

Null hypothesis:\mu \leq 13.2  

Alternative hypothesis:\mu > 13.2  

The statistic for this case is:

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

Calculating the statistic

Replacing the info given we got:

t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436    

P-value

The degrees of freedom are:

df=n-1=44-1=43  

Since is a right tailed test the p value is:  

p_v =P(t_{(43)}>1.436)=0.079  

Decision

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

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