If you can help me solve this, I\'d really appreciate it! Willrate all answers :

Question

If you can help me solve this, I'd really appreciate it! Willrate all answers :) Suppose the average annual spending ratio on education for asample of 25 countries is normally distributed with a mean of 32%and standard deviation of 14%. Calculate a 95% confidence intervalfor the average annual spending ratio on education for allcountries. Interpret your interval. If you can help me solve this, I'd really appreciate it! Willrate all answers :) Suppose the average annual spending ratio on education for asample of 25 countries is normally distributed with a mean of 32%and standard deviation of 14%. Calculate a 95% confidence intervalfor the average annual spending ratio on education for allcountries. Interpret your interval.

Explanation / Answer

mean , = 32 standard deviation , = 14 standard error, e = / n = 14/25 =2.8 at 95% confidence, for a two tail distribution, z =1.96             (from z table) a 95% confidence interval for the average annual spendingratio on education for all countries = ± (z*e) = 32 ± (1.96*2.8) = (26.512, 37.488) The above interval says average annual spendingratio on education will lie between 26.512% and 37.488% at a 95%confidence.
The above interval says average annual spendingratio on education will lie between 26.512% and 37.488% at a 95%confidence.

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