Show intro/Instructions According to the Bureau of Labor Statistics, the mean sa

Question

Show intro/Instructions According to the Bureau of Labor Statistics, the mean salary for registered nurses in Kentucky was $56,277. The distribution of salaries is assumed to be normally distributed with a standard deviation of $5,872. Someone would like to determine if registered nurses in Ohio have a greater average pay. To investigate this claim, a sample of 199 registered nurses is selected from the Ohio Board of Nursing, and each is asked their annual salary. The mean salary for this sample of 199 nurses is found to be $56,702.512. a. Completely describe the sampling distribution of the sample mean salary when samples of size 199 are selected. (round your answer to 4 decimal places) standard deviation: = shape: the distribution ofy is o select an answer B because select an answer o b. What conjecture has been made? The mean salary for registered nurses in Ohio is $56,277. The mean salary for registered nurses in Ohio is greater than S56,277. The mean salary for registered nurses in Ohio is $56,702.512. The mean salary for registered nurses in Ohio is greater than S56,702.512. c. Using the distribution described in part a, what is the probability of observing a sample mean of 56,702.512 or more? (round to 2 decimal places) o probability- (include 4 decimal places) d. Based on the probability found, what conclusion can be reached? o The probability would be classified as Select an answer.So, there Select an answer evidence to support the conjecture that the mean salary for registered nurses in Ohio is greater than S56,277.

Explanation / Answer

Solution:

We are given

Population mean = µ = 56277

Population standard deviation = ? = 5872

Sample size = n = 199

Sample mean = Xbar = 56702.512

Part a

Mean of sampling distribution: µybar = µ = 56277

(Mean of the sampling distribution is equal to the population mean.)

Standard deviation: ?ybar = ?/sqrt(n) = 5872/sqrt(199) = 416.2550

(Standard deviation of the sampling distribution is equal to the standard error ?/sqrt(n) of the population mean.)

The distribution of ybar is approximately normal because sample size is adequately large and we know that the sampling distribution of sample statistic approaches to the normal as we increase the sample size.

Part b

For the given scenario, following conjecture has been made:

The mean salary for registered nurses in Ohio is greater than $56277.

Part c

Here, we have to find P(Xbar ? 56702.512)

P(Xbar ? 56702.512) = P(Xbar > 56702.512)

P(Xbar ? 56702.512) = 1 - P(Xbar < 56702.512)

Z = (Xbar - µ) / [?/sqrt(n)]

Z = (56702.512 – 56277) / [5872/sqrt(199)]

Z = (56702.512 - 56277) /416.2550

Z = 425.512/416.2550

Z = 1.022239

Z = 1.02

P(Z<1.02) = 0.846666

P(Xbar<56702.512) = 0.846666

P(Xbar ? 56702.512) = 1 - P(Xbar < 56702.512)

P(Xbar ? 56702.512) = 1 - 0.846666

P(Xbar ? 56702.512) = 0.153334

Required probability = 0.1533

Part d

We get probability (P-value) as 0.1533 which is greater than 0.05, so it is not significant.

The probability would be classified as non-significant. So, there is insufficient evidence to support the conjecture that the mean salary for registered nurses in Ohio is greater than $56277.

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