Question

Suppose that we know the average production of peanuts in the state of Virginia is 3000 pounds per acre. A new plant food has been developed and is tested on 61 individual plots of land. The mean yield with the plant food is 3120 pounds per acre, with a standard deviation of 578 pounds per acre. Since the plant food is moderately expensive, we will only recommend using it if we can show it increases peanut production at a 5% significance level.

The alternative hypothesis for this test is : U>3000

The distribution (choose from Z, T, X^2, or F) for this test is 1.62

The critical value for this test is .0526

The test value for this test is _______

Based on this information, we (choose reject or fail to reject)   the null hypothesis

Based on this we (choose from do or do not)   recommend using this plant food.

Answer 1

1)The alternative hypothesis for this test is : µ >3000

2)The distribution for this test is :t (with n-1 =60 degree of freedom)

3)0.05 level with right tail test and n-1= 60 df, critical t=

1.671

4)

population mean μ=		3000
sample mean 'x̄=		3120.000
sample size    n=		61
std deviation s=		578.0000
std error ='sx=s/√n=578/√61=		74.0053
t statistic ='(x̄-μ)/sx=(3120-3000)/74.005=		1.622

The test value for this test is=1.622

Based on this information, we fail to reject

Based on this we do not recommend using this plant food.