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dezoksy [38]
2 years ago
12

Venn diagrams in maths

Mathematics
1 answer:
OleMash [197]2 years ago
5 0

Answer:

Step-by-step explanation:

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The amount of time it takes a bat to eat a frog was recorded for each bat in a random sample of 12 bats. The resulting sample me
spin [16.1K]

Answer: a. CI for the mean: 17.327 < μ < 26.473

b. CI for variance: 29.7532 ≤ \sigma^{2} ≤ 170.9093

Step-by-step explanation:

a. To construct a 95% confidence interval for the mean:

The given data are:

mean = 21.9

s = 7.7

n = 12

df = 12 - 1 = 11

1 - α = 0.05

\frac{\alpha}{2} = 0.025

t-score = t_{0.025,11} = 2.2001

Note: since the sample population is less than 30, it is used a t-score.

The formula for interval:

mean ± t.\frac{s}{\sqrt{n} }

Substituing values:

21.9 ± 2.200.\frac{7.7}{\sqrt{12} }

21.9 ± 4.573

The interval is: 17.327 < μ < 26.473

b. A 95% confidence interval for the variance:

The given values are:

s^{2} = 7.7^{2}

s^{2} = 59.29

α = 0.05

\frac{\alpha}{2} = 0.025

1-\frac{\alpha}{2} = 0.975

\chi^{2}_{0.025,11} = 21.92

\chi^{2}_{0.975,11} = 3.816

Note: To find the values for \chi^{2}_{\alpha/2,n-1} and \chi^{2}_{1-\alpha/2,n-1}, look for them at the chi-square table

The formula to calculate interval:

(\frac{(n-1).s^{2}}{\chi^{2}_{\alpha/2,n-1}} , \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2,n-1}})

are the lower and upper limits, respectively.

Substituing values:

(\frac{11.59.29}{21.92} , \frac{11.59.29}{3.816})

(29.7532, 170.9093)

The interval for variance is: 29.7532 ≤ \sigma^{2} ≤ 170.9093

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Compounded Interest. Compounded quarterly.
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