In: Statistics and Probability
A). Suppose Travel and Leisure reported the average hotel price in Miami, Florida, was $153.57 per night in 2019. Assume the population standard deviation is $26.86 and that a random sample of 30 hotels was selected. Calculate the standard error of the mean.
B). According to the US Labor Department, the average hourly wage for private-sector production and non-supervisory workers was $20.04 in February 2013. Assume the standard deviation for this population is $6.00 per hour. A random sample of 35 workers from this group was selected. What is the standard error of the mean?
C). According to the US Labor Department, the average hourly wage for private-sector production and non-supervisory workers was $20.04 in February 2013. Assume the standard deviation for this population is $6.00 per hour. A random sample of 35 workers from this group was selected. What is the probability that the mean for this sample is less than $19.00?
D). According to the US Labor Department, the average hourly wage for private-sector production and non-supervisory workers was $20.04 in February 2013. Assume the standard deviation for this population is $6.00 per hour. A random sample of 35 workers from this group was selected. What is the probability that the mean for this sample is more than $20.84??
How would we interpret the probability calculated in the questions D?
E). According to the US Labor Department, the average hourly wage for private-sector production and non-supervisory workers was $20.04 in February 2013. Assume the standard deviation for this population is $6.00 per hour. A random sample of 35 workers from this group was selected. What is the probability that the mean for this sample is exactly $20.00?
A)
The standard error of the mean is 4.9039.
B)
The standard error of the mean is 1.0142.
C)
D)
E)
Since normal distribution is continuous distribution so probability at a specific point is zero.
So the probability that the mean for this sample is exactly $20.00 is 0.