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Question 7

Hiroyuki surveyed 325 randomly chosen students asking how much time they spent getting ready for school in the morning. The average time spent was 26
minutes with a standard deviation of 5.8 minutes.
Use a 90% confidence interval to find the maximum error of the estimate of time spent getting ready for school. Round to the nearest hundredth
We can say with 90% confidence that the mean time spent getting ready for school is within
minutes of the sample mean time.

1 Answer

2 votes
To find the maximum error of the estimate of time spent getting ready for school, we can use the formula:

Maximum error = z* (standard deviation / square root of sample size)

where z is the z-score corresponding to the desired level of confidence. For a 90% confidence interval, z is 1.645.

Substituting the given values, we get:

Maximum error = 1.645 * (5.8 / sqrt(325))
Maximum error = 0.66 (rounded to two decimal places)

Therefore, the maximum error of the estimate of time spent getting ready for school is 0.66 minutes.

To find the 90% confidence interval for the mean time spent getting ready for school, we can use the formula:

Confidence interval = sample mean +/- z* (standard deviation / square root of sample size)

Substituting the given values, we get:

Confidence interval = 26 +/- 1.645 * (5.8 / sqrt(325))
Confidence interval = 26 +/- 0.66
Confidence interval = (25.34, 26.66)

Therefore, we can say with 90% confidence that the mean time spent getting ready for school is within 0.66 minutes of the sample mean time, or between 25.34 and 26.66 minutes.
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