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shtirl [24]
3 years ago
11

In a random sample of 8 people, the mean commute time to work was 36.5 minutes and the standard deviation was 7.4 minutes. A 98%

confidence interval using the t-distribution was calculated to be (28.7,44.3). After researching commute times to work, it was found that the population standard deviation is 8.7 minutes. Find the margin of error and construct a 98% confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare the results.The margin of error of u is ___A 98% confidence interval using the standard normal distribution is (__,__)Compare the results. Choose the correct answer below.a. the confidence interval found using the standard normal distribution has smaller lower and upper confidence interval limitsb. the confidence interval found using the standard normal distribution is wider than the confidence interval found using the students t-distributionc. the confidence interval found using the standard noraml distribution is the same as the confidence interval found using the students t-distributiond. the confidence interval found using the standard normal distrubution is narrower than the confidence interval found using the students t-distribution
Mathematics
1 answer:
ANEK [815]3 years ago
3 0

Answer:

Step-by-step explanation:

Given that

Sample size = n = 8: sample mean = 36.5 and s = sample std dev = 7.4 minutes

98% CI using t = (28.7, 44.3) with margin of error as 7.8

Now if sigma = population std dev is known we use Z critical value

Margin of error = 2.33*8.7/\sqrt 8 = 7.167

CI = (36.5 ±7.167)

This is not wider than that used using t distribution

The margin of error of u is ___7.167

98% confidence interval using the standard normal distribution is (29.333__,43.667__)

d)the confidence interval found using the standard normal distrubution is narrower than the confidence interval found using the students t-distribution

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