hospital recruiting company for doctors is conducting a study to determine whether the average doctor's salary (in thousands of dollars) is significantly more than $89 thousand dollars. A random sample of 26 doctors' salaries (in thousands of dollars) is given in the table. Test the hospital recruiting company's claim using a 5% level of significance. Salary (in thousands of dollars) 86.04 95.33 79.51 98.13 83.51 93.58 95.87 90.17 94.34 85.21 88.77 85.35 91.72 94.62 95.27 86.85 103.31 94.33 92.9 90.78 90.36 96.03 92.85 92.27 95.83 93.51 Preliminary: Is it safe to assume that n≤5%n≤5% of all doctors earning a salary? Yes No Is n≥30n≥30 ? No Yes Since n<30n<30 , then determine whether the box plot of salaries contains any outliers. 7580859095100105Salary[Graphs generated by this script: setBorder(15); initPicture(75,105,-3,4);axes(5,100,1,null,null,1,'off');text([86.9,-3],"Salary");line([79.51,2],[79.51,4]); rect([88.77,2],[95.27,4]); line([92.875,2],[92.875,4]);line([103.31,2],[103.31,4]); line([79.51,3],[88.77,3]); line([95.27,3],[103.31,3]);] There are outliers from the box plot. There aren't outliers from the box plot. Note: Without being given that the population is normally distributed, and given a sample less than 30, we are required to verify that there are no outliers by plotting a box plot and a normal probability plot. Test the claim: What are the null and alternative hypotheses? H0:H0: Select an answer σ s μ s² x̄ σ² p̂ p ? = < > ≠ ≥ ≤ H1:H1: Select an answer s p̂ σ² s² μ p x̄ σ ? < ≠ > = ≤ ≥

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Description: A hospital recruiting company for doctors is conducting a study to determine whether the average doctor's salary (in thousands of dollars) is significantly more than $89 thousand dollars. A random sample of 26 doctors' salaries (in thousands of dolla

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ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

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A hospital recruiting company for doctors is conducting a study to determine whether the average doctor's salary (in thousands of dollars) is significantly more than $89 thousand dollars. A random sample of 26 doctors' salaries (in thousands of dollars) is given in the table. Test the hospital recruiting company's claim using a 5% level of significance.

Salary (in thousands of dollars)

86.04	95.33	79.51	98.13	83.51
93.58	95.87	90.17	94.34	85.21
88.77	85.35	91.72	94.62	95.27
86.85	103.31	94.33	92.9	90.78
90.36	96.03	92.85	92.27	95.83
93.51

Preliminary:

Is it safe to assume that n≤5%n≤5% of all doctors earning a salary?
- Yes
- No
Is n≥30n≥30 ?
- No
- Yes
Since n<30n<30 , then determine whether the box plot of salaries contains any outliers.

7580859095100105Salary[Graphs generated by this script: setBorder(15); initPicture(75,105,-3,4);axes(5,100,1,null,null,1,'off');text([86.9,-3],"Salary");line([79.51,2],[79.51,4]); rect([88.77,2],[95.27,4]); line([92.875,2],[92.875,4]);line([103.31,2],[103.31,4]); line([79.51,3],[88.77,3]); line([95.27,3],[103.31,3]);]
- There are outliers from the box plot.
- There aren't outliers from the box plot.

Note: Without being given that the population is normally distributed, and given a sample less than 30, we are required to verify that there are no outliers by plotting a box plot and a normal probability plot.

Test the claim:

What are the null and alternative hypotheses?

H0:H0: Select an answer σ s μ s² x̄ σ² p̂ p ? = < > ≠ ≥ ≤
H1:H1: Select an answer s p̂ σ² s² μ p x̄ σ ? < ≠ > = ≤ ≥
Based on the hypotheses, calculate the following. Round to four decimal places.
Test Statistic =
pp-value =
Make a decision.
Select an answer Fail to reject the null hypothesis Reject the null hypothesis Accept the null hypothesis Accept the alternative hypotheis
Make a conclusion regarding the claim.

Select an answer There is enough evidence to reject the claim There is enough evidence to support the claim There is not enough evidence to reject the claim There is not enough evidence to support the claim that the population mean doctor's salary (in thousands of dollars) is significantly more than $89 thousand dollars.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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