nancytrumblesty
Answered

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A study of 36 camels showed that they could walk an average rate of 4.2 kilometers per hour. The sample standard deviation is 0.64. Find the 90% confidence interval of the mean for all camels. ​

Sagot :

Answer:

HI BRAINLY USER!

A study of 36 camels showed that they could walk an average rate of 4.2 kilometers per hour. The sample standard deviation is 0.64. Find the 90% confidence interval of the mean for all camels

90% CI for using Normal dist

Sample Mean=7=4.2

Since sample size is large, sample standard deviation can be taken as an approx-

imation of population standard deviation.

Population Standard deviation =⁰ = 0.64

Sample Size = n = 36

Significance level = a = 1- Confidence 1-0.9 = 0.1

The Critical Value = a/2 = 20.05 = 1.645 (From z table, using interpolation,1/2 th distance between 1.64 and 1.65 )

Critical Values, a/2 = ±1.645

Margin of Error, E= 0.175)

Limits of 90% confidence interval are given by:

90% confidence interval is: E=4.2±0.175467

= (4.024533, 4.375467)

90% CI using normal dist: 4.025 < p < 4.375

Also, the exact answer using TI-83/84 or excel is

4.2+ invNorm(0.95) x 0.64/sqrt(36)

= (4.0245483982.4.3754516018)

Note that some professor prefer 1-distribution when standard deviation is un known even though sample size is more than 30, in this case the answer will be: 90% CI using t-dist: 4.020 < p < 4.380

3. If the standard deviation of the lifetimes of television tubes is estimated as 100 hours, how large a sample must be selected in order to be 90% confident that the error in the estimated mean lifetime will not exceed 20 hours?

exceed 20 hours.

at least 68

10 hours

ANS: (b) at least 271

Step-by-step explanation:

QUESTION

A study of 36 camels showed that they could walk an average rate of 4.2 kilometers per hour. The sample standard deviation is 0.64. Find the 90% confidence interval of the mean for all camels

___________________________________

[tex]ANSWER & SOLUTION[/tex]

90% CI for using Normal dist

Sample Mean=7=4.2

Since sample size is large, sample standard deviation can be taken as an approx

imation of population standard deviation.

Population Standard deviation = ⁰ = 0.64

  • Sample Size = n = 36
  • Significance level = a = 1- Confidence 1 0.9 = 0.1
  • The Critical Value = a/2 = 20.05 = 1.645

(From z table, using interpolation,1/2 th distance

between 1.64 and 1.65 )

Critical Values, a/2 = ±1.645

[tex]Marginof Error E = za/2 \times = \sqrt[0]{n} = 1.645 = \sqrt[0.64]{36} ≈0.175467[/tex]

Margin of Error, E= 0.175)

Limits of 90% confidence interval are given by:

[tex]Lowerlimit -E-4.2 - 0.1754674.0245334.025[/tex]

[tex]Upperlimit=7+ E= 4.2+ 0.1754674.3754674.375[/tex]

  • 90% confidence interval is: E=4.2+0.175467
  • = (4.024533, 4.375467)

90% CI using normal dist: 4.025 < p < 4.375

Also, the exact answer using TI-83/84 or excel is

  • ➝ 4.2+ invNorm(0.95) x 0.64/sqrt(36)
  • = (4.0245483982.4.3754516018)

__________________________________

Note : that some professor prefer 1 distribution when standard deviation is un known even though sample size is more than 30, in this case the answer will be: 90% CI using t-dist: 4.020 < p < 4.380

#CARRY ON LEARNING

– zhen

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