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Molodets [167]
2 years ago
15

Help please

Mathematics
1 answer:
kifflom [539]2 years ago
4 0

Answer:

1. false

2. false

3. true

Step-by-step explanation:

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Determining Whether a Difference Is Statistically
SCORPION-xisa [38]

Using the z-distribution, it is found that:

  • The 95% confidence interval is of -1.38 to 1.38.
  • The value of the sample mean difference is of 1.74, which falls outside the 95% confidence interval.

<h3>What is the z-distribution confidence interval?</h3>

The confidence interval is:

\overline{x} \pm zs

In which:

  • \overline{x} is the difference between the population means.
  • s is the standard error.

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.

The estimate and the standard error are given by:

\overline{x} = 0, s = 0.69

Hence the bounds of the interval are given by:

\overline{x} - zs = 0 - 1.96(0.69) = -1.38

\overline{x} + zs = 0 + 1.96(0.69) = 1.38

1.74 is outside the interval, hence:

  • The 95% confidence interval is of -1.38 to 1.38.
  • The value of the sample mean difference is of 1.74, which falls outside the 95% confidence interval.

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

#SPJ2

6 0
2 years ago
Which of the following expressions results in 0 when evaluated at x = 4?
Zarrin [17]
(x-10)(x-4) because 4-10=6 then multiply that by 0 from 4-4
8 0
3 years ago
The table displays the frequency of scores for a calculus class on an exam. The mean of the exam scores is 3.5. Score 1 2 3 4 5
vladimir1956 [14]

Answer:

f=11

Mode=4

Median=4

Explanation:

We are given that

a.Mean of the exam score,\bar x=3.5

Score(x)   frequency   C.F

1                  1                 1

2                  3                4

3                  f                 15(4+f=4+11)

4                  13               28

5                   4               32

\sum f_i=1+3+f+13+4=21+f

\sum f_ix_i=1(1)+2(3)+3(f)+4(13)+5(4)=1+6+3f+52+20=79+3f

(\bar x)=\frac{\sum f_ix_i}{\sum f_i}

Using the formula

3.5=\frac{79+3f}{21+f}

73.5+3.5f=79+3f

3.5f-3f=79-73.5

0.5f=5.5

f=\frac{5.5}{0.5}=11

b.Mode:The number which is repeat most times .

4 repeat most times

Hence, mode of all exam scores=4

N=32

N is even

Median=\frac{(\frac{n}{2})^{th}+(\frac{n}{2}+1)^{th}}{2}

Median=\frac{16th+17th}{2}=\frac{4+4}{2}=\frac{8}{2}=4

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