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Fudgin [204]
3 years ago
9

In order to test whether the average waiting time this year differs from last year, a sample of 25 data were collected this year

with a mean of 0.74 hours and standard deviation 0.22 hours. Calculate the 99% confidence interval for the true waiting time for banking service this year.
a. 0.74 ± 2.57 x 0.22/√25
b. 0.74 ± 2.797 x 0.22/√25
c. 0.74 ± 2.797 x 0.22/√24
d. 0.74 ± 2.787 x 0.22/√25
Mathematics
2 answers:
DerKrebs [107]3 years ago
6 0

Answer:

a. 0.74 ± 2.57 x 0.22/√25

Step-by-step explanation:

Confidence interval is a range of values in which there is a specified probability that the value of a parameter lies within that range.

The confidence interval of a statistical data can be written as.

x+/-zr/√n .......1

Given;

Mean gain x = 0.74 hours

Standard deviation r = 0.22

Number of samples n = 25

Confidence interval = 99%

z (at 99% confidence) = 2.57 (from the table)

Substituting the values we have into equation 1;

0.74+/- 2.57×0.22/√25

netineya [11]3 years ago
4 0

Answer:

(B) 0.74 + or - 2.797 × 0.22/√25

Step-by-step explanation:

Confidence Interval = mean + or - t × sd/√n

mean = 0.74 hours

sd = 0.22 hours

n = 25

degree of freedom = n - 1 = 25 - 1 = 24

Confidence level = 99%

t-value corresponding to 24 degrees of freedom and 99% confidence level is 2.797

Confidence Interval = 0.74 + or - 2.797 × 0.22/√25

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Answer:

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Step-by-step explanation:

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3 years ago
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Si en 11 cajas de chocolates hay 220 chocolates en total. Calcula cuántos chocolates habrá en total en 5 cajas de chocolates.
pochemuha

Answer:

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Step-by-step explanation:

Esta pregunta es resolvida usando proporciones, con lo uso de regla de tres .

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