A different sample of size n = 20 produced a sample mean of 31.08% and a sample standard deviation of 5.48%.

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Question:

A different sample of size n = 20 produced a sample mean of 31.08% and a sample standard deviation of 5.48%. Use these values to calculate a 90% confidence interval for the national mean percent of low-income working families. Please show all calculations and provide the upper and lower limits that make up the confidence interval. (Round the limits to two decimal places).

A random sample of size n = 20 which produced a sample mean of 31.08% and sample standard deviation of 5.48%.

Thus  n = 20    , \bar{x}=30.08 , s = 5.48

We have to find 90% confidence interval for the national mean percent of low income working families.

Since sample size  = n = 20 is small and population standard deviation = is unknown.

Thus we use t distribution to find confidence interval.

Formula for confidence interval for mean is :

(\bar{x}-E\: , \: \bar{x}+E)