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ipn [44]
3 years ago
10

Inferential statistical tests are used for all of the following except: a. describing a sample with means and standard deviation

s b. determining if results occurred by chance c. testing hypotheses by asking if there are differences between groups d. making assumptions about a population
Mathematics
1 answer:
Anika [276]3 years ago
8 0

Answer:

a. Describing a sample with mean and standard deviation.

Step-by-step explanation:

Statistics can be categorized into descriptive and inferential statistics.

descriptive statistics uses data for descriptions through numerical analysis. It can be further divided in four parts.

  • Measures of Central Tendency ( Mean, Median, and Mode)
  • Measures of Frequency (Count, Percent, Frequency)
  • Measures of Position (Percentile Ranks, Quartile Ranks.)
  • Measures of Dispersion ( Range, Standard Deviation)

Inferential statistics however is based on assumptions and conclusions and generalizations drawn from samples or checks.

options b to d are all examples of inferential statistics while option a is an example of descriptive statistics.

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During science class,jimmy heated a substance and recorded the temperature of the substance over several minutes. He used the da
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It’s -18 that’s the answer

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3 years ago
Recall that a 6-bit string is a bit strings of length 6, and a bit string of weight 3, say, is one with exactly three 1's. How m
strojnjashka [21]

Answer:

1.. Total number of 6 bit strings is 64

2. Number of 6-bit strings with weight of 0 is 1

3. Number of 6-bit strings with weight of 1 is 6

4. Number of 6-bit strings with weight of 3 is 20

5. Number of 6-bit strings with weight of 5 is 6

6. Number of 6-bit strings with weight of 6 is 1

7. Number of 6-bit strings with weight of 7 is 0

Step-by-step explanation:

A bit string is a string that contains 0 and 1 only

1. Total number of 6 bit strings is 2^6 = 64

2. Number of 6 bit strings with weight 0 is 1

Explanation

Weight 0 means a string with no occurrence of 1

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 0 i.e. no occurrence of 1

6C0 = 6!/(6-0)!0!

6C0 = 6!/6!0!

6C0 = 1

Hence, the number of string with weight 0 (i.e. no occurrence of 1 ) is 1

3. Number of string with weight 1 is 6

Explanation

Weight 0 means a string with exactly 1 occurrence of '1'

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 1

6C1 = 6!/(6-1)!1!

6C1 = 6!/5!1!

6C1 = 6

Hence, the number of string with weight 6

4. Number of string with weight 3 is 20

Explanation

n = 6 and r = 3

6C3 = 6!/(6-3)!3!

6C3 = 6!/3!3!

6C3 = 20

Hence, the number of string with weight 3 is 20

5. Number of string with weight 5 is 6

Explanation

n = 6 and r = 5

6C5 = 6!/(6-5)!5!

6C5 = 6!/1!5!

6C5 = 6

Hence, the number of string with weight 5 is 6

6. Number of string with weight 6 is 1

Explanation

n = 6 and r = 6

6C6 = 6!/(6-6)!6!

6C6 = 6!/0!6!

6C6 = 1

Hence, the number of string with weight 6 is 1

7. Number of string with weight 7 is 0

Weight of 7 means that a string that has 7 occurrence of 1

The total length of a 6 bit is 6

Since 6 is less than 7, there's no way a bit of weight 7 can occur.

So, the right answer for this is 0.

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answer:

43 FT^2

Step-by-step explanation:

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