question archive Suppose a survey of 912 cars listed for sale in Florida in 2020 finds an average price of 13,80 with a standard deviation of 3,140 construct a 90% interval for the average price what is the lower end of the confidence interval round to the nearest car
Suppose a survey of 912 cars listed for sale in Florida in 2020 finds an average price of 13,80 with a standard deviation of 3,140 construct a 90% interval for the average price what is the lower end of the confidence interval round to the nearest car
Subject:StatisticsPrice:2.86 Bought3
Suppose a survey of 912 cars listed for sale in Florida in 2020 finds an average price of 13,80 with a standard deviation of 3,140 construct a 90% interval for the average price
what is the lower end of the confidence interval round to the nearest car
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Values are given as,
Mean, x bar = 13800,
Sample Standard Deviation, s = 3140,
Sample Size, n = 912,
Degree of freedom, df = n - 1 = 911,
alpha = 1 - (90/100) = 0.100,
Using df = 911 and alpha = 0.100 in t Table,
t-critical value = 1.647
Calculating lower bound of 90% confidence interval:
Now, using the formula
?x?−t∗(s/n?)?
Put the values in the above formula:
?13800−1.647∗(3140/912?)?
= 13800 -1.647*103.9758
= 13800 - 171.2481
= 13628.7519
= 13629
The lower endpoint of 90% CI = [13629]
