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Musya8 [376]
2 years ago
7

Am I right or wrong?? ​

Mathematics
1 answer:
Slav-nsk [51]2 years ago
8 0

Answer:

i think thats correct

Step-by-step explanation:

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Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation: 22, 18, 1
BigorU [14]

Answer:

a:  Sample mean: 20, Sample standard deviation: 1.732

b:  19.13 <µ < 20.87

c:  18.84< µ < 21.16

d:  as the confidence level increases, the interval becomes wider

Step-by-step explanation:

a:  Sample mean is found by adding up all the individual values of the sample, then dividing by the total number of values in the sample.  

Sample standard deviation is the square root of the variance

See the first attached photo for the calculations of these values.

We need to create a 80% confidence interval for the population.  Since n < 30, we will use a t-value with degree of freedom of 7 (the degree of freedom is always one less than the sample size.  

Look for the column on the t-distribution chart that has "area in two tails" of 0.20 (80%), and row 7 (degree of freedom)

The t-value is 1.415

See the second attached photo for the construction of the confidence interval

We need to create a 90% confidence interval for the population.  Since n < 30, we will use a t-value with degree of freedom of 7 (the degree of freedom is always one less than the sample size.  

Look for the column on the t-distribution chart that has "area in two tails" of 0.10 (90%), and row 7 (degree of freedom)

The t-value is 1.895

See the third attached photo for the construction of the confidence interval

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3 years ago
(Fractions) Rewrite the expression using fithteenths. 2/5 - 1/3 =
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Answer:

1/15

Step-by-step explanation:

2/5-1/3=6/15-5/15=1/15

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Case bought a 10 week old pig that weighed 30 kg. At 18 weeks the pig weighed 80 kg. The equation (18 − 10)w + 30 = 80 can be us
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Step-by-step explanation:

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