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masya89 [10]
3 years ago
13

Which of the following statistical measures is least appropriate for describing the data set shown below?

Mathematics
1 answer:
stealth61 [152]3 years ago
4 0

mean is the answer nnnnnnnnn

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Help please I don’t get itttttttt I’ll mark brainlist if I can pleaseee
Marizza181 [45]

Answer:

<u>$5.94 </u>

Step-by-step explanation:

Multiply 2.75 by $0.76 to get $2.09

Multiply 2.75 by $1.40 to get $3.85

Add $3.85 and $2.09 to get $5.94

6 0
3 years ago
Day care centers expose children to a wider variety of germs than the children would be exposed to if they stayed at home more o
Ostrovityanka [42]

Answer:

a) The study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is approximately 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is approximately 0.8565

c) The odds ratio is approximately 0.6780

d) The 95% CI is approximately 0.523 < OR < 0.833

e) i) Yes

ii) Children with more social activity have a higher occurrence of acute lymphoblastic leukemia

Step-by-step explanation:

a) An experimental study is a study in which the researcher adds inputs to the study and then monitors the outcomes

An observational study is one in which the researcher observes risk factors and does not intervene in the process

Therefore, the study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is \hat p_1 = 1020/1272 = 85/106 = 0.8\overline {0188679245283} ≈ 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is \hat p_2 = 5343/6238 ≈ 0.8565

c) We have the following two way table;

\begin{array}{ccc}Lymphoblastic \ Leukemia &With \ Social \ Activity & Without \ Social \ Activity\\With&1020 \ (a)&252 \ (b)\\Without&5343 \ (c)&895 \ (d) \end{array}

The \ odds \ ratio = \dfrac{1,020 \times 895}{5,343 \times 252} \approx 0.6780

The odds ratio ≈ 0.6780

d) The 95% confidence interval for the odds ratio is given as follows;

95 \% \ CI = OR \ \pm \ 1.96 \times \sqrt{\dfrac{1}{a} +\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}}

Where;

a = 1020, b = 252, c = 5343, d = 895

Therefore;

95 \% \ CI \approx  0.6780 \ \pm \ 1.96 \times \sqrt{\dfrac{1}{1,020} +\dfrac{1}{252}+\dfrac{1}{5,343}+\dfrac{1}{895}} \approx 0.678 \pm0.155

The 95% CI ≈ 0.523 < OR < 0.833

e) i) Given that the observed Odds Ratio is within the Confidence Interval, therefore, there is an indication that the amount of social activity is associated with acute lymphoblastic leukemia

ii) Based on the proportion of the study findings, children with more social activity have a higher occurrence of acute lymphoblastic leukemia

6 0
3 years ago
Why is 2/3 greater than 3/5? Explain this to an 8 year old
Dmitrij [34]

\text{Hey there!}

\bf{\frac{2}{3}=0.6\overline{6}7}\\\\ \bf{\frac{3}{5}=0.60}

\text{Reason: if you use a fraction pie chart and compare}\frac{2}{3} \text{to}\frac{3}{5}, \text{you'd see that}\frac{2}{3} \text{has more shaded areas than as for}\frac{3}{5}\text{has the least amount}

\text{Good luck on your assignment and enjoy your day!}

~\frak{LoveYourselfFirst:)}

8 0
3 years ago
-7 x (4+8) = ?? Please a aswer?
Colt1911 [192]
The answer would be -84.

Here are the steps I did.

1. Simplify 4 + 8 to 12 ( -7 * 12 )

2. Now lastly simplify and u get -84.
5 0
3 years ago
Read 2 more answers
Divide j by 9 and the multiply the quotient by 2
zhenek [66]

Answer:

Step-by-step explanation:

you will need to do:

j/9*2=2j/9

3 0
3 years ago
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