1. A simple random sample of 400 students is taken at a large university. The average height of the sampled students is 68 inches and the SD is 2 inches. The distribution of heights in the sample follows the normal curve very closely.
An approximate 68% confidence for the average height of students at the university is 68 inches plus or minus ____________ inches.
Approximately 68% of the students in the sample have heights in the range 68 inches plus or minus ____________ inches.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.34) and its Z score (score in terms of SD).
A. 68% = mean ± 1 SEm
SEm = SD/√n
B. 68% = mean ± 1 SD
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A simple random sample of voters is taken from the voters in a large state. Using the methods of our course, researchers construct an approximate 99% confidence interval for the percent of the state’s voters who will vote for Candidate A. The interval goes from 37.3% to 48.7%.
In the sample, the percent of voters who will vote for Candidate A
is equal to ____________%.
An approximate 95% confidence interval for the percent of the state’s voters who will vote for Candidate A goes from _________% to _________%.
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A simple random sample of voters is taken from the voters in a large state. Using the methods of our course, researchers construct an approximate 99% confidence interval for the percent of the state’s voters who will vote for Candidate A. The interval goes from 37.3% to 48.7%.
In the sample, the percent of voters who will vote for Candidate A
is equal to ____________%.
An approximate 95% confidence interval for the percent of the state’s voters who will vote for Candidate A goes from _________% to _________%.
To find the approximate 68% confidence interval for the average height of students at the university, we can use the standard error of the mean.
First, we need to calculate the standard error of the mean (SEM) using the formula:
SEM = SD / √(sample size)
= 2 / √(400)
= 2 / 20
= 0.1 inches
Next, we can determine the margin of error (MOE) by multiplying the SEM by the z-score corresponding to the desired confidence level. For a 68% confidence level, the z-score is approximately 1.
MOE = SEM * z-score
= 0.1 * 1
= 0.1 inches
Therefore, the approximate 68% confidence interval for the average height of students at the university is 68 inches plus or minus 0.1 inches.
Now, to find the range within which approximately 68% of the students in the sample have heights, we can use the same margin of error (MOE) and subtract it from and add it to the sample mean.
Lower bound = Sample mean - MOE
= 68 - 0.1
= 67.9 inches
Upper bound = Sample mean + MOE
= 68 + 0.1
= 68.1 inches
Approximately 68% of the students in the sample have heights in the range of 67.9 inches to 68.1 inches.