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suppose that a random sample of 41 state college students is asked to measure the length of their right foot in centimeters. an 85% confidence interval for the mean foot length for students at this university turns out to be (21.709, 25.091). if instead a 90% confidence interval was calculated, how would it differ from the 85% confidence interval?

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The 90% Confidence interval, is become narrow than the 85% confidence interval.

Confidence Intervals:

Confidence intervals are the range of estimates for unknown parameters. Confidence intervals are computed at the specified confidence level.

C.I = X-Bar ± Z(S ÷ √n)

where,

X Bar ---> Sample mean

Z --> Confidence level

S ---> Sample standard deviation

n ---> Sample size

We have given that,

Sample Size (n) = 41

Confidence Level = 85%

Confidence Level We assumed the 85% confidence interval to be (21.709, 25.091).

We asked if we calculated an 85% confidence interval instead, how does that differ from a 90% confidence interval?

Let's assume this is a 90% confidence interval. All other conditions like sample size or other are the same. when we calculate a 90% confidence interval, its narrower the interval and we are less confident about true population parameter lies within that interval, and the more we trust that the population parameter lies within a wider interval, it narrows. So, at 5% confidence, we can have very tight intervals.

To learn more about Confidence interval, refer:

https://brainly.com/question/26658887

#SPJ4.

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