# Need help as quickly as possible???

#### penneymmp

Question -
A government program that was designed to address poverty established guidelines by which cities could procure special grant money for development in any area that has an average ncome significantly below the Minimal Cost of Living Index. Researchers identified the poorest area, drew a random sample of 62 families, and asked for annual income information. They found a mean family income of $9637.00, with a satndard deviation of$420.00. Given an Index of $9724.00, will they receive the grant? How do you statistically solve the problem? Can you post the steps? Thanks. #### curquht ##### New Member Question - A government program that was designed to address poverty established guidelines by which cities could procure special grant money for development in any area that has an average ncome significantly below the Minimal Cost of Living Index. Researchers identified the poorest area, drew a random sample of 62 families, and asked for annual income information. They found a mean family income of$9637.00, with a satndard deviation of $420.00. Given an Index of$9724.00, will they receive the grant?

How do you statistically solve the problem? Can you post the steps?

Thanks.
Sounds like a one sample t problem to me. You use a t-distribution when the population standard deviation is unknown.
I always start off the problem writing down everything i know:

n = 62
x_bar <- $9637.00 s <-$420.00
x <- $9724.00 you have everything you need to compute the t score: t= (x-x_bar)/(s/sqrt(n)) When looking up a t-score in your t-disbribution table, you also have to know the degrees of freedom. They are n-1. The t-score will generally be between two values, so that means that your probability will be between two p-values. Hope this helps! #### Dr.D ##### New Member Hi penneymmp, It is not based on the t-distribution but you have to do a Z-test since n > 30. It is the formula: (Sample Mean - Population Mean)/s/sqrt(n) Sample mean (x_bar) =$9637.00
Population mean = $9724.00 Standard deviation = s =$420.00
N = 62

When you find the z-value, you will use 5% level of significance, the z-tables should show you a critical value of 1.96 on both sides. If the calculated value is within -1.96 to +1.96, you say that the mean is not significantly lower than the minimum index, and they should not get the grant. But it is outside the range, then it is significantly lower than the index and they are entitled to the grant.