Question 1051565: From a random sample of 70 high school seniors in a particular school district , the mean and standard deviation of the verbal scores in the scholastic Aptitute Test (SAT) are found to be 433 and 47, respectively. Based on this sample, construct a 98 % confidence interval for the mean verbal score in the SAT for the population of all seniors in this school district.
Thank you
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! Our first step is to calculate the Margin of Error(ME)
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Since we know the standard deviation of the statistic, we use the following
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ME = critical value * standard deviation of the statistic
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Since our sample size is > 30, we use the normal distribution and our critical
value is the z-score
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Next, calculate alpha
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alpha(a) = 1 - (confidence level / 100) = 0.02
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critical probability(p^*) = 1 - (a/2) = 0.99
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the z-score associate with p^* is 2.32
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Therefore ME = 2.33 * 47 = 109.51
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Our 98% confidence interval(CI) = sample statistic + or - ME
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CI = 433 + or - 109.51 = (323.49, 542.51)
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