Question

SOLVE ONLY PART E

has mean that has changecl tom 40 Assume that, when 1. There is a concern a voltage generator tested, the output χ will have mean μ with standard deviation σ2) HoY 40 vs. Ha:μ#40 a) (3pts) What kinds of values of 3 would lead to the rejection of the null hypothesis? In other words, what is the correct form of the rejection region (RR) ? SELECT A,B,C,D,E, or F: Answer A) Significantly larger than 40 B) larger than 40 C) Significantly smaller than 40 F) different than 40 D) smaller than 40 E) Significantly different than 40 b) Suppose a sample of size 64 produces a sample mean (average) of 39.78 Find the p-value. For each of c) (4 pts) Suppose a new sample is taken, and the p-value is computed using the test above. the two (hypothetical) p-values below, decide which is the best interpretation of the result. (Assume significance level is 0.05) p-value of 0.1156 p-value of 0.0015 i) the mean voltage has decreased from 40 ii) the mean voltage has changed from 40 i) the mean voltage may or may not have changed from 40 iv) the mean voltage has increased from 40 v) the mean voltage equals 40 exactly. 40.5 (3 pts) Suppose that, in reality but unbeknownst to the researcher, the population meanise What would be the result if the researcher does not reject the null hypothesis? ) type l error (i) type Il error il) type Illerror iv) a correct decisionf answer HO.S e) Suppose that, in reality but unbeknownst to the researcher, the population mean is what is the probability the researcher does not reject the null hypothesis, assuming the significance level a0.10 is adopted, and the sample size is 100? 40

Answer 1

e)

true mean ,    µ =    40.5                          
                                  
hypothesis mean,   µo =    40                          
significance level,   α =    0.1                          
sample size,   n =   100                          
std dev,   σ =    2                          
                                  
δ=   µ - µo =    0.5                          
                                  
std error of mean,   σx = σ/√n =    2   / √    100   =   0.2000          
                                  
Zα/2   = ±   1.645   (two tailed test)                      
We will fail to reject the null (commit a Type II error) if we get a Z statistic between                           -1.645   and   1.645
these Z-critical value corresponds to some X critical values ( X critical), such that                                  
                                  
-1.645   ≤(x̄ - µo)/σx≤   1.645                          
39.671   ≤ x̄ ≤   40.329                          
                                  
now, type II error is ,ß =        P (   39.671   ≤ x̄ ≤   40.329   )          
       Z =    (x̄-true mean)/σx                      
       Z1 = (   39.671   -   40.5   ) /   0.2000   =   -4.145
       Z2 = (   40.329   -   40.5   ) /   0.2000   =   -0.855
                                  
   so, P(   -4.145   ≤ Z ≤   -0.855   ) = P ( Z ≤   -0.855   ) - P ( Z ≤   -4.145   )
                                  
       =   0.196   -   0.000   =   0.1962 (answer) [ Excel function: =NORMSDIST(z) ]