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In a random sample of 8 ​people, the mean commute time to work was 33.5 minutes and the standard deviation was 7.2 minutes. A 95​% confidence interval using the​ t-distribution was calculated to be (27.5 comma 39.5 ). After researching commute times to​ work, it was found that the population standard deviation is 9.5 minutes. Find the margin of error and construct a 95​% confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare the results.

Respuesta :

Answer: Margin of error = 6.58

Confidence interval =  (26.91, 40.08)

Step-by-step explanation:

Since we have given that

Sample size n = 8

Sample mean = 33.5 minutes

Population standard deviation = 9.5 minutes

At 95% confidence interval,

α = 0.05

t = 1.96

So, Margin of error is given by

[tex]t\times \dfrac{\sigma}{\sqrt{n}}\\\\=1.96\times \dfrac{9.5}{\sqrt{8}}\\\\=6.58[/tex]

Confidence interval would be

Lower limit:

[tex]\bar{x}-6.58\\\\=33.5-6.58\\\\=26.91[/tex]

Upper limit:

[tex]\bar{x}+6.58\\\\=33.5+6.58\\\\=40.08[/tex]

Hence, the interval would be (26.91, 40.08)

But at standard deviation 7.2 minutes, the confidence interval was (27.5, 39.5)

Confidence interval using the standard normal distribution is wider than the confidence interval using t distribution.

The information regarding the statistics shows that the margin of error is 6.58.

How to calculate the margin of error

From the information given, the following can be deduced:

  • Sample size = 8
  • Standard deviation of population = 9.5
  • Sample mean = 33.5

Therefore, the margin of error will be:

= 1.96 × 9.5/✓8

= 6.58

Also, the lower limit will be:

= 33.5 - 6.58

= 26.91

The upper limit will be:

= 33.5 + 6.58

= 40.08

Therefore, the interval will be (26.91, 40.08)

When the standard deviation is 7.2 minutes, the confidence interval is (27.5, 39.5).

Learn more about margin of error on:

https://brainly.com/question/10218601

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