Question 1: We want to estimate the mean change in score µ in the population of...

Question 1: We want to estimate the mean change in score µ in the population of all high school seniors. An SRS of 450 high school seniors gained an average of x⎯⎯⎯x¯ = 21 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation 52.201.

Find σx¯, the standard deviation of the mean change x¯ _______ (±±0.001).

Using the 68-95-99.7 Rule (Empirical Rule), give a 95% confidence interval for μμ based on this sample.

Confidence interval (±±0.001) is between _____ and _______.

Question 3: We have the survey data on the body mass index (BMI) of 670 young women. The mean BMI in the sample was x¯=25.3. We treated these data as an SRS from a Normally distributed population with a standard deviation σ=7.8.

Give confidence intervals for the mean BMI and the margins of error for 90%, 95%, and 99% confidence.

Conf. Level	Interval (±±0.01)	margins of error (±±0.0001)
90%	______ to _____	_______
95%	_____ to ______	_______
99%	_____ to ______	_______

Solutions

Expert Solution

Find σx¯, the standard deviation of the mean change

Answer)

Standard deviation for mean = s.d/√n = 52.201/√450. = 2.461

Using the 68-95-99.7 Rule (Empirical Rule), give a 95% confidence interval for μμ based on this sample.

Confidence interval (±±0.001)

According to the emperical rule

If the data is normally distributed

Then 68% lies in between mean - s.d and mean + s.d

95% lies in between mean - 2*s.d and mean + 2*s.d

99.7% lies in between mean - 3*s.d and mean + 3*s.d

So, 95% lies in between 21 - (2*2.461) and 21 + (2*2.461)

That is in between 16.078 and 25.922

orchestra answered 1 year ago

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