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A medical resident is convinced that, on average, residents at his hospitals work for over 3 hours more per week than residents at other hospitals. Suppose that he finds that a random sample of 30 residents at his hospital works an average of 74.3 hours per week with a standard deviation of 2.6 hours. He also surveys a random sample of 30 residents working at other hospitals and finds they worked an average of 70.1 hours per week with a standard deviation of 2.9 hours. Do the data support the resident’s belief? Test the relevant hypotheses at a significance level of 0.05.

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Answer:

At 0.05 significance level, there is significant evidence that residents at the medical resident's hospitals work for over 3 hours more per week than residents at other hospitals.

Step-by-step explanation:

[tex]H_{0}[/tex]:residents at the medical resident's hospitals work for 3 hours more per week than residents at other hospitals

[tex]H_{a}[/tex]:residents at the medical resident's hospitals work for over 3 hours more per week than residents at other hospitals.

Test statistic can be calculated using the formula:

z=[tex]\frac{(X-Y)-3}{\sqrt{\frac{s(x)^2}{N(x)}+\frac{s(y)^2}{N(y)}}}[/tex] where

  • X is the average working hours at the medical resident's hospitals (74.3)
  • Y is the average working hours at other hospitals (70.1)
  • s(x) is the sample standard deviation of working hours of the resident's hospitals (2.6)
  • s(y) is the sample standard deviation of working hours of the other hospitals (2.9)
  • N(x) is the sample size of working hours of the resident's hospitals (30)
  • N(y) is the sample size of working hours of the other hospitals(30)

z=[tex]\frac{(74.3-70.1)-3}{\sqrt{\frac{2.6^2}{30}+\frac{2.9^2}{30}}}[/tex] ≈ 1.6875

P-value of the test statistic is   0.046. Since  0.046>0.05 we reject the null hypothesis.

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