the mean amount of money spent per week on gas by a sample of 25 drivers was found to be $57.00 with a standard deviation of $2.36. assuming the population distribution is normally distributed construct and interpret a 90% confidence interval for the mean amount of money spent on Gas per week

The sample mean is 57, the standard deviation is 2.36, n s 25 and z is the upper (1-C)/2 critical value for the standard normal distribution. Here, as we want a confidence interval at a 90% we have (1-C)/2=0.05 we have to look at the 1-0.05=0.95 value at the normal distribution table, which is 1.65 approximately. Replacing all these values:

ci = 57 +- 1.65*(2.36)/[25^(1/2)]

ci = 57 +- 3.894*/(5) = 57 +- 0.78

ci => (56.22 , 57.78)

the mean amount of money spent per week on gas by a sample of 25 drivers was found to be $57.00 with a standard deviation of $2.36. assuming the population distribution is normally distributed construct and interpret a 90% confidence interval for the mean amount of money spent on Gas per week​

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the mean amount of money spent per week on gas by a sample of 25 drivers was found to be $57.00 with a standard deviation of $2.36. assuming the population distribution is normally distributed construct and interpret a 90% confidence interval for the mean amount of money spent on Gas per week