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jasenka [17]
2 years ago
9

Researchers were interested in whether relaxation training decreases the number of headaches a person experiences. They randomly

assigned 20 participants to either a control group or a relaxation training group and recorded the change in number of headaches each group reported the week before training to the week after training (the control group received no training). Which statistical analysis would we perform to see whether the training helped the participants in that group as compared to the control group
Mathematics
1 answer:
wolverine [178]2 years ago
3 0
A line graph or a chart table
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The obtained answers for the given frequency distribution are:

(a) The formula for the mean in sigma notation is \bar x =\frac{1}{n}  \sum X_i where n is the number of observations; X_i are the n observations.

The mean for the given monthly plan price is $16.1.

(b) The frequency distribution for given data is {$9.99 - 2; $10 - 5; $12 - 1; $12.75 - 2; $14.99 - 6; $20 - 4; $25 - 5}

(c) The formula for the mean using the frequency distribution table is \bar x = \frac{1}{N}\sum f_ix_i where N =\sum f_i and on applying this formula for the given data, the mean is $16.1.

(d) The median for the given data is m_e = 14.99, and the mode for the given data is $14.99

<h3>What are the mean, median, and mode for a frequency distribution?</h3>

The frequency distribution has sample observations x_i and frequencies f_i.

Then, the mean is calculated by

\bar x = \frac{1}{N}\sum f_ix_i

Where N =\sum f_i (Sum of frequencies)

The median is calculated by

m_e=\left \{ {{x_{k}} \ if \ n = 2k+1 \atop {\frac{x_{k}+x_{k+1}}{2}} \ if \ n =2k} \right.

The mode is calculated by

Mode = highest frequency value

<h3>Calculation:</h3>

The given list of data is

{$14.99, $12.75, $14.99, $14.99, $9.99, $25, $25, $10, $14.99, $10, $20, $10, $20, $14.99, $10, $25, $20, $12, $14.99, $25, $25, $20, $12.75, $10, $9.99}

(a) Formula for the mean using sigma notation and use it to calculate the mean:

The formula for the mean is

\bar x =\frac{1}{n}  \sum X_i

Where n = 25; X_i - n observations

On substituting,

Mean \bar x

=1/25(14.99+12.75+14.99+14.99+9.99+25+25+10+14.99+10+20+10+20+14.99+10+25+20+12+14.99+25+25+20+12.75+10+9.99)

= 1/25(402.42)

= 16.09 ≅ 16.1

(b) Constructing a frequency distribution for the data:

Cost - frequency - cumulative frequency

$9.99 - 2 - 2

$10 - 5 - 7

$12 - 1 - 8

$12.75 - 2 - 10

$14.99 - 6 - 16

$20 - 4 - 20

$25 - 5 - 25

Sum of frequencies N = 25;

(c) Using frequency distribution, calculating the mean:

The formula for finding the mean using frequency distribution is

\bar x = \frac{1}{N}\sum f_ix_i

Where N = 25;

On substituting,

\bar x<em> </em>= 1/25 (2 × 9.99 + 5 × 10 + 1 × 12 + 2 × 12.75 + 6 × 14.99 + 4 × 20 + 5 × 25)

  = 1/25 (402.42)

  = 16.09 ≅ 16.1

Therefore, the mean is the same as the mean obtained in option (a).

(d) Calculating the median and the mode:

Since N = 25(odd) i.e., 2· 12 + 1; k = (12 + 1)th term = 13th term

So,  the median m_e = 14.99. (frequency at 13th term)

Since the highest frequency is 6 occurred by the cost is $14.99,

Mode = 14.99

Learn more about frequency distribution here:

brainly.com/question/27820465

#SPJ9

7 0
2 years ago
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