n = 36  H_{0}: μ ≤ 20  
= 24.6  H_{a}: μ > 20  
σ = 12 
1
Thirty six GGC students were sampled to determine their average age. The GGC website indicates that the average age of students was 20 years old, and you decided to conduct an upper tail test because you think the average age is higher. You collected the above information. What is the pvalue of your hypothesis test.
Question options:
0.5107 

0.0214 

0.0107 

2.1 
2
n = 36  H_{0}: μ ≤ 20  
= 24.6  H_{a}: μ > 20  
σ = 12 
Thirty six GGC students were sampled to determine their average age. The GGC website indicates that the average age of students was 20 years old, and you decided to conduct an upper tail test because you think the average age is higher. You collected the above information. Based on your analysis, the null hypothesis should ___
Question options:
not be rejected 

be rejected 

Not enough information is given to answer this question. 

None of the other answers are correct. 
3
You sampled 100 voters. Eighty percent of the people in the sample liked the incumbent. We are interested in determining whether or not the proportion of the population in favor of the incumbent is significantly more than the hypothesized 75%.
Question options:
2.00 

1.25 

0.05 

0.80 
4
You sampled 100 voters. Eighty percent of the people in the sample liked the incumbent. We are interested in determining whether or not the proportion of the population in favor of the incumbent is significantly more than the hypothesized 75%. The pvalue for this hypothesis is
Question options:
0.2112 

0.0156 

0.025 

0.05 
Ans 1 ) using minitab>stat>basic stat>one sample z
we have
OneSample Z
Test of μ = 20 vs > 20
The assumed standard deviation = 12
N Mean SE Mean 95% Lower Bound Z P
36 24.60 2.00 21.31 2.30 0.0107
P value = 0.0107
2
since p value 0.0107 <0.05 so null hypothesis should be
rejected
Ans 3 using minitab>stat>basic stat>one sample proportion 'we have
Test and CI for One Proportion
Test of p = 0.75 vs p ≠ 0.75
Sample X N Sample p 95% CI ZValue PValue
1 80 100 0.800000 (0.721601, 0.878399) 1.15 0.248
Using the normal approximation.
z = 1.15
4 p value is 0.248
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