A survey was conducted to see how much people spend on gas. In a random sample of 15 drivers, the average weekly gas expense was $57 with a standard deviation of $2.36. Assuming the amount of money drivers spend on gas is normally distributed, compute a 95% confidence interval for the mean gas cost of all drivers. Round ALL answers to 3 decimal places. To compute this CI, I will need to use the distribution (z or t). The critical value (from the table) that I need to compute the margin of error is: . The margin of error for my interval is: . The lower bound of my CI is: and the upper bound of my Ci is: .
A survey was conducted to see how much people spend on gas. In a random sample of 15 drivers, the average weekly gas expense was $57 with a standard deviation of $2.36. Assuming the amount of money drivers spend on gas is
To compute this CI, I will need to use the distribution (z or t).
The critical value (from the table) that I need to compute the margin of error is: .
The margin of error for my interval is: .
The lower bound of my CI is: and the upper bound of my Ci is: .
95% confidence interval:
1-α =1 – 0.95 = 0.05 [Using Excel formula “=NORM.S.INV (1-(0.05/2))”]
Therefore, the critical value is obtained is 1.960
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