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March 6, 2015

March 6, 2015

Posted by **looking for someone to varify my work please** on Tuesday, April 13, 2010 at 10:39pm.

a. Use the critical value z0 method from the normal distribution.

Answer:

1. H0 : µ = .35

Ha : µ ≠ .35

2.a = 0.05 (one tail)

3.Test statistics: (0.44-0.35)/ (0.21/√20) = 1.92

4.P-value or critical z0 or t0.: z0 = 1.96

5. Rejection Region: Z<-1.96 or Z>1.96

6. Decision: do not reject Ho

7. Interpretation: not sufficient evidence to conclude that the mean level of pesticide is greater that the limit of 0.35 ppm.

b. Use the P-value method.

Answer:

1.H0 :

Ha :

2.a =

3.Test statistics:

4.P-value or critical z0 or t0.

5.Rejection Region:

6.Decision:

7.Interpretation:

- statistics -
**PsyDAG**, Wednesday, April 14, 2010 at 1:58pma. If you are using a one-tailed test, then:

Ha: μ > .35

You would reject, if Z ≥ 1.645 for P ≤ .05.

In your conclusion, you need to state that the value is or is not "*significantly*greater."

I did not check your calculations.

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