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Gnom [1K]
3 years ago
5

The mean is the measure of central tendency most likely to be affected by an outlier.

Mathematics
1 answer:
rjkz [21]3 years ago
4 0

Answer:

The correct option is A) The statement is true.

Step-by-step explanation:

Consider the provided statement.

The Mean is calculated as: sum of values / number of values

Outlier is the value that "lies outside" most of the other values in a set of data. Or the values that are much smaller or larger.

The mean is the measure of central tendency.

Now consider the provided statement.

The statement says mean is measure of central tendency which is true.

It also say that it is affected by an outlier.

This is true because mean is the sum of values divided by the number of values. Also it is most affected by the extreme values .

Thus, the correct option is A) The statement is true.

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Hello people can you help me with dis two question
____ [38]

Answer

length = 11 in

width = 6in

2 in = 5m

length =

28m ÷ 5

= 5

remainder 3m

1 in = 2.5

if 2in = 5m

then 2 × 5 = 10

10 + 1 = 11in

width

2in = 5m

15m ÷ 5

= 3

3× 2

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3 0
2 years ago
There are big spenders among University of Alabama football season ticket holders. This data set Roll Tide!! shows the dollar am
Bas_tet [7]

Using the z-distribution, as we have a proportion, the 95% confidence interval is (0.2316, 0.3112).

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which:

  • \pi is the sample proportion.
  • z is the critical value.
  • n is the sample size.

In this problem, we have a 95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so the critical value is z = 1.96.

We also consider that 130 out of the 479 season ticket holders spent $1000 or more at the previous two home football games, hence:

n = 479, \pi = \frac{130}{479} = 0.2714

Hence the bounds of the interval are found as follows:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2714 - 1.96\sqrt{\frac{0.2714(0.7286)}{479}} = 0.2316

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2714 + 1.96\sqrt{\frac{0.2714(0.7286)}{479}} = 0.3112

The 95% confidence interval is (0.2316, 0.3112).

More can be learned about the z-distribution at brainly.com/question/25890103

7 0
2 years ago
Can someone help me with this??
Shtirlitz [24]

Answer:

here you go

Step-by-step explanation:

18 is 24c/d^5

19 is 48a^/b^3

/ indicates a fraction

^ indicates an exponent

7 0
3 years ago
A survey of 1,562 randomly selected adults showed that 522 of them have heard of a new electronic reader. The accompanying techn
tester [92]

Answer:

a) We want to test the claim that 35​% of adults have heard of the new electronic reader, then the system of hypothesis are.:  

Null hypothesis:p=0.35  

Alternative hypothesis:p \neq 0.35  

And is a two tailed test

b) z=\frac{0.334 -0.35}{\sqrt{\frac{0.35(1-0.35)}{1562}}}=-1.326  

c) p_v =2*P(z  

d) Null hypothesis:p=0.35  

e) Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha.

Step-by-step explanation:

Information provided

n=1562 represent the random sample selected

X=522 represent the people who have heard of a new electronic reader

\hat p=\frac{522}{1562}=0.334 estimated proportion of people who have heard of a new electronic reader

p_o=0.35 is the value to verify

\alpha=0.05 represent the significance level

z would represent the statistic

p_v represent the p value

Part a

We want to test the claim that 35​% of adults have heard of the new electronic reader, then the system of hypothesis are.:  

Null hypothesis:p=0.35  

Alternative hypothesis:p \neq 0.35  

And is a two tailed test

Part b

The statistic for this case is given :

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

Replacing the info given we got:

z=\frac{0.334 -0.35}{\sqrt{\frac{0.35(1-0.35)}{1562}}}=-1.326  

Part c

We can calculate the p value using the laternative hypothesis with the following probability:

p_v =2*P(z  

Part d

The null hypothesis for this case would be:

Null hypothesis:p=0.35  

Part e

The best conclusion for this case would be:

Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha.

5 0
3 years ago
Need Help ASAP……………..
deff fn [24]

Answer:

Okay, but doesn't this bring any trouble reading?

4 0
2 years ago
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