In a certain city , the hourly wage of workers on temporary employment crontracts is normally distributed. The mean is $15 and t
1 answer:
Answer:
16%
Step-by-step explanation:
The mean is $15 and the standar deviation is $3.
mean = 15 and SD = 3
We need to find percentage less than 12 per hour
P(x<12)= P(x=12)
to find P(x=12) we find z-score
Now use z-score table . z-score = 0.1587
P(x=12)=0.1587
To get percentage we multiply by 100
0.1587 * 100 = 15.87 = 16%
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<h2><u>Answer with explanation</u>:</h2>
Let be the population mean.
As per given , we have
Since the alternative hypothesis is right-tailed , so the test is a right-tailed test.
Also, population standard deviation is given , so we perform one-tailed z-test.
Test statistic :
, where = Population mean
= Population standard deviation
n= sample size
= Sample mean
For n= 18 , ,
,
, we have
P-value (for right tailed test): P(z>2.12) = 1-P(z≤ 2.12) [∵ P(Z>z)=1-P(Z≤z)]\
=1- 0.0340=0.9660
Decision : Since P-value(0.9660) > Significance level (0.01), it means we are failed to reject the null hypothesis.
[We reject null hypothesis if p-value is larger than the significance level . ]
Conclusion : We do not have sufficient evidence to show that the goal is not being met at α = .01 .