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Construct a 90% confidence interval for the population mean, . assume the population has a normal distribution. a sample of 15 randomly selected students has a grade point average of 2.86 with a standard deviation of 0.78

User Leo Ribeiro
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1 Answer

19 votes
19 votes

Answer:


CI=\{2.5053,3.2147\}

Explanation:

Assuming the conditions for constructing a t-confidence interval are true, we use the formula
\displaystyle CI=\bar{x}\pm t^*\biggr((s)/(√(n))\biggr) where our sample mean is
\bar{x}=2.86, our sample standard deviation is
s=0.78, our sample size is
n=15, and a 90% confidence level is equivalent to a critical value of
t^*=1.7613,


\displaystyle CI=\bar{x}\pm z^*\biggr((s)/(√(n))\biggr)\\\\CI=2.86\pm 1.761\biggr((0.78)/(√(15))\biggr)\\\\CI=2.86\pm0.3547\\\\CI=\{2.5053,3.2147\}

Hence, we are 90% confident that the true population mean of a student's grade point average is between 2.5053 and 3.2147

User Grimsock
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