Hello everyone TIA 1 Find the Pvalue for the indicated hypot

Hello everyone, TIA

1- Find the P-value for the indicated hypothesis test with teh given standardized test statistic, z. Decide whether to reject H o for the given level of signifiance a.                     Right tailed test with test statistic z=1.92 and a=0.02

P-value = _ (Round to four decimal places as needed. 1

2. Assume the random variable x is normally distributed with mean u=50 and standard deviation o=7. Find the indicated probability. P(x>41) =_       Round to four decimal places as needed.

3. Assume the random variable x is normally distributed with mean u=81 and standard deviation o=4. Find the indicated probability.   P (68<x<73) =_    round to four decimal places as needed.

4. Find the margin of error for the given values of c, s and n.   C=0.90, S=3.3, N=49   E=_

5.Find the margin of error for the given values of C,S and n. C=0.99, s=5, n=25. The margin of error is _ Round to one decimal place as needed?

6. Find the margin of error for the given values of c,s and n.   C=0.95, S=2.4, N=15?

Solution

1.

The right tailed area for z = 1.92 is

P(z > 1.92) = 0.02742895 [ANSWER, P VALUE]

Thus, this P > 0.02.

We only reject Ho when P < significance level.

Thus, WE FAIL TO REJECT HO.

2.

          
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    41      
u = mean =    50      
n = sample size =    1   [just one random variable]  
s = standard deviation =    7      
          
Thus,          
          
z =    -1.285714286      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -1.285714286   ) =    0.900728603

3.

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    68      
x2 = upper bound =    73      
u = mean =    81      
n = sample size =    1      
s = standard deviation =    4      
          
Thus, the two z scores are          
          
z1 = lower z score =    -3.25      
z2 = upper z score =    -2      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.000577025      
P(z < z2) =    0.022750132      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.022173107      

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