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ioda
2 years ago
9

A chi-square test for goodness of fit is used to examine the distribution of individuals across three categories, and a chi-squa

re test for independence is used to examine the distribution of individuals in a 2×3 matrix of categories. Which test has the larger value for df?
Mathematics
1 answer:
aleksandrvk [35]2 years ago
3 0

Answer:

Equal df

Step-by-step explanation:

Given that a chi square test for goodness of fit is used to examine the distribution of individuals across three categories,

Hence degree of freedom = 3-1 =2

Similarly for a chi-square test for independence is used to examine the distribution of individuals in a 2×3 matrix of categories.

Here degree of freedom = (r-1)(c-1) where r = no of rows and c =no of columns

= (2-1)(3-1) = 2

Thus we find both have equal degrees of freedom.

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Monica is multiplying 8.714 by 1,000. How many places should she move the decimal point?
Rom4ik [11]

How should you write the proportion 9:36 = 10:40 using words?  

     

  A. 9 is to 36 as 40 is to 10  

  B. 9 is to 10 as 36 is to 40  

  C. 36 is to 9 as 10 is to 40  

  D. 9 is to 36 as 10 is to 40

7 0
2 years ago
A store has a $120 dress. Then there is a 200% increase in the price of the dress. What is the final price of the dress?
LenKa [72]

Multiply 120 by 200 percent and you would get 240, then you would add 240 and 120 to get 360.

7 0
2 years ago
If LaTeX: m\angle ABF=8s-6m ∠ A B F = 8 s − 6 and LaTeX: m\angle ABE=2\left(s+11\right)m ∠ A B E = 2 ( s + 11 ), find LaTeX: m\a
kolbaska11 [484]

Answer:

<h2><em>2(3s-14)</em></h2>

Step-by-step explanation:

Given the angles ∠ABF=8s-6, ∠ABE = 2(s + 11), we are to find the angle ∠EBF. The following expression is true for the three angles;

∠ABF = ∠ABE + ∠EBF

Substituting the given angles into the equation to get the unknown;

8s-6 = 2(s + 11)+ ∠EBF

open the parenthesis

8s-6 = 2s + 22+ ∠EBF

∠EBF = 8s-6-2s-22

collect the like terms

∠EBF = 8s-2s-22-6

∠EBF = 6s-28

factor out the common multiple

∠EBF = 2(3s-14)

<em></em>

<em>Hence the measure of angle ∠EBF is 2(3s-14)</em>

8 0
2 years ago
J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.
melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

                      P.Q.  =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }  ~ t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean for mail-order sales = $74.50

\bar X_2 = sample mean for internet sales = $84.40

s_1 = sample standard deviation for mail-order purchases = $17.25

s_2 = sample standard deviation for internet purchases = $21.25

n_1 = sample of sales receipts for mail-order purchases = 16

n_2 = sample of sales receipts for internet purchases = 9

Also,  s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }  =  \sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} } = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by (\mu_1-\mu_2).

Now, 99% confidence interval for (\mu_1-\mu_2) is given by;

             = (\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_)  \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

          = (74.50-84.40) \pm (2.807  \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})

          = [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0
2 years ago
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2 years ago
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