Question 1191082: A researcher is interested in finding a 90% confidence interval for the mean number of times per day that college students text. The study included 100 students who averaged 37.6 texts per day. The standard deviation was 24.5 texts. Round answers to 3 decimal places where possible.
a. To compute the confidence interval use a
?
distribution.
b. With 90% confidence the population mean number of texts per day is between
and
texts.
c. If many groups of 100 randomly selected members are studied, then a different confidence interval would be produced from each group. About
percent of these confidence intervals will contain the true population number of texts per day and about
percent will not contain the true population mean number of texts per day.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! n = 100
sm = 37.6
ssd = 24.5
se = 24.5/sqrt(100) = 2.45
since the mean and standard deviation are from the sample, use the t-distribution.
at 90% two-tailed confidence interval, the critical t-score, with 99 degrees of freedom, will be equal to plus or minus 1.660 rounded to 2 decimal places.
t-score formula is:
t = (x - m) / se.
when t = -1.660, the t-score formula becomes:
-1.660 = (x - 37.6) / 2.45.
solve for x to get:
x = -1.660 * 2.45 + 37.6 = 33.533.
when t = 1.660, the t-score formula becomes:
1.660 = (x - 37.6) / 2.45.
solve for x to get:
x = 1.660 * 2.45 + 37.6 = 41.667
at 90% confidence interval, the sample mean will be between 33.533 and 41.667.
if many samples of 100 elements each were selected from the population, it is estimated that approximately 90% of these samples will contain the true population mean within the confidence limits and 10% of these samples intervals will not contain the true population mean within the confidence limits.
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