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olganol [36]
2 years ago
7

HELP AND EXPLAIN

Mathematics
1 answer:
omeli [17]2 years ago
3 0

The relative frequency of 15/24 for this considered situation represents the situation given by: Option C: Out of all the family members that prefer a lake, 15 are children.

<h3>How to find the relative frequency?</h3>

Relative frequency is the ratio of the considered sub group's count to the total count. (so its frequency of the considered sub group relative to the total frequency). (subgroup being group defined by a row, or a column)

<h3>How to form two-way table?</h3>

Suppose two dimensions are there, viz X and Y. Some values of X are there as X_1, X_2, ... , X_n and some values of Y are there as Y_1, Y_2, ... , Y_n

List them in title of the rows and left to the columns. There will be n \times k table of values will be formed(excluding titles and totals), such that:

Value(ith row, jth column) = Frequency for intersection of X_i and Y_j (assuming X values are going in rows, and Y values are listed in columns).

Then totals for rows, columns, and whole table are written on bottom and right margin of the final table.

For n = 2, and k = 2, the table would look like:

\begin{array}{cccc}&Y_1&Y_2&\rm Total\\X_1&n(X_1 \cap Y_1)&n(X_1\cap Y_2)&n(X_1)\\X_2&n(X_2 \cap Y_1)&n(X_2 \cap Y_2)&n(X_2)\\\rm Total & n(Y_1) & n(Y_2) & S \end{array}

where S denotes total of totals, also called total frequency.

n is showing the frequency of the bracketed quantity, and intersection sign in between is showing occurrence of both the categories together.

The given two-way frequency table is:

                                        Lake                  Hike                       Total

Children                                                       6

Adults                                 9

Total                                                             14                            38

Total of hike subgroup is 14, out of which children are 6, so adults going for hiking must be 14 - 6 = 8 as adults and children going for hiking together are 14 in counts.

So, we get:
n(Hike ∩ Children) = 8

Thus, the total of subgroup Adults is 9+8=17 = n(Adults)

Total of vertical total columns is: n(Children) + n(Adults) = 38

or, we get: n(Children) = 38 - 17=  21

Since total of children subgroup is:

n(Lake ∩ Children) + n(Hike ∩ Children) = n(Children)

Thus, we get:
n(Lake ∩ Children) + 6 = 21

n(Lake ∩ Children) = 21-6 = 15

Thus, sum of the 'lake' subgroup is:

n(Lake) = n( Lake ∩ Children) + n(Lake ∩ Adults)

n(Lake) = 15 + 9 = 24

Thus, the completed two-way table is:

                                        Lake                  Hike                       Total

Children                             15                        6                             21

Adults                                 9                         8                             17

Total                                   24                       14                            38

Now, we've to find what does 15/24 represents.

The division is done by 24, which is the total of "lake" subgroup.

And in that lake subgroup, we have:

n(Lake ∩ Children) = 15

Thus, 15/24 is the relative frequency of those children who prefer the lake, relative to total number of people who prefer lake.

Thus, the relative frequency of 15/24 for this considered situation represents the situation given by: Option C: Out of all the family members that prefer a lake, 15 are children.

Learn more about conditional relative frequency here:

brainly.com/question/8358304

Learn more about two-way table here:

brainly.com/question/26788374

#SPJ1

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Answer:

Lower = 231.134- 3.098=228.036

Upper = 231.134+ 3.098=234.232

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n=5 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And for this case we know that the 95% confidence interval is given by:

\bar X=\frac{233.002 +229.266}{2}= 231.134

And the margin of error is given by:

ME = \frac{233.002 -229.266}{2}= 1.868

And the margin of error is given by:

ME= t_{\alpha/2} \frac{s}{\sqrt{n}}

The degrees of freedom are given by:

df = n-1 = 5-1=4

And the critical value for 95% of confidence is t_{\alpha/2}= 2.776

So then we can find the deviation like this:

s = \frac{ME \sqrt{n}}{t_{\alpha/2}}

s = \frac{1.868* \sqrt{5}}{2.776}= 1.506

And for the 99% confidence the critical value is: t_{\alpha/2}= 4.604

And the margin of error would be:

ME = 4.604 *\frac{1.506}{\sqrt{5}}= 3.098

And the interval is given by:

Lower = 231.134- 3.098=228.036

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Answer:

(a) A 95% confidence interval for the population mean is [433.36 , 448.64].

(b) A 95% upper confidence bound for the population mean is 448.64.

Step-by-step explanation:

We are given that article contained the following observations on degrees of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:

420, 425, 427, 427, 432, 433, 434, 437, 439, 446, 447, 448, 453, 454, 465, 469.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                              P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean = \frac{\sum X}{n} = 441

            s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }  = 14.34

            n = sample size = 16

            \mu = population mean

<em>Here for constructing a 95% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.</em>

<em />

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-2.131 < t_1_5 < 2.131) = 0.95  {As the critical value of t at 15 degrees of

                                             freedom are -2.131 & 2.131 with P = 2.5%}  

P(-2.131 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.131) = 0.95

P( -2.131 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.131 \times {\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-2.131 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.131 \times {\frac{s}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X -2.131 \times {\frac{s}{\sqrt{n} } } , \bar X +2.131 \times {\frac{s}{\sqrt{n} } } ]

                                      = [ 441-2.131 \times {\frac{14.34}{\sqrt{16} } } , 441+2.131 \times {\frac{14.34}{\sqrt{16} } } ]

                                      = [433.36 , 448.64]

(a) Therefore, a 95% confidence interval for the population mean is [433.36 , 448.64].

The interpretation of the above interval is that we are 95% confident that the population mean will lie between 433.36 and 448.64.

(b) A 95% upper confidence bound for the population mean is 448.64 which means that we are 95% confident that the population mean will not be more than 448.64.

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