Question

Salary 77600 76000 90700 97200 90700 101800 78700 81300 84200 97600 77500 75700 89400 84300 78700 84600 87700 103400 83800 101300 94700 69200 95400 61500 68800 The national mean annual salary for a school administrator is $90,000 a year (The Cincinnati Enquirer, April 7, 2012). A school official took a sample of 25 school administrators in the state of Ohio to learn about salaries in that state and to see if they differed from the national average. Click on the datafile logo to reference the data. a. Formulate hypotheses that can be used to determine whether the population mean annual administrator salaries in Ohio differ from the national mean of $90,000. Null hypothesis: 1. 2. 3. Choose correct answer from above choice Alternative hypothesis: 1. 2. 3. Choose correct answer from above choice b. Use the sample data for the 25 Ohio administrators and the t distribution table (Table 2 in Appendix B) to estimate a p-value range for your hypothesis test in part (a). The p-value is c. At = .05, can your null hypothesis be rejected? What is your conclusion? p-value .05, H 0. We that the mean annual administrator salary in Ohio the national mean annual salary. d. Repeat the preceding hypothesis test using the critical value approach. Round your answer to two decimal places. Enter negative values as negative numbers. t = ; H 0

Answer #1

**Solution:**

Given: sample of 25 observations.

A school official took a sample of 25 school administrators in the state of Ohio to learn about salaries in that state and to see if they differed from the national average.

Part a) Formulate hypotheses that can be used to determine whether the population mean annual administrator salaries in Ohio differ from the national mean of $90,000.

vs

Part b) Use the sample data for the 25 Ohio administrators and the t distribution table (Table 2 in Appendix B) to estimate a p-value range for your hypothesis test in part (a).

Find t test statistic:

where

Thus we need to make following table:

x Salary | (x - xbar)^2 |

77600 | 58859584 |

76000 | 85969984 |

90700 | 29463184 |

97200 | 142277184 |

90700 | 29463184 |

101800 | 273174784 |

78700 | 43191184 |

81300 | 15776784 |

84200 | 1149184 |

97600 | 151979584 |

77500 | 60403984 |

75700 | 91623184 |

89400 | 17040384 |

84300 | 944784 |

78700 | 43191184 |

84600 | 451584 |

87700 | 5895184 |

103400 | 328624384 |

83800 | 2166784 |

101300 | 256896784 |

94700 | 88887184 |

69200 | 258309184 |

95400 | 102576384 |

61500 | 565107984 |

68800 | 271326784 |

Thus t test statistic :

For p-value , find df = n -1 = 25 - 1 = 24

Now look in t table for df = 24 row and find an interval in which t = 2.141 fall.

and then find corresponding two tail area.

t = 2.141 fall in between 2.064 and 2.492 and corresponding two tail area is in between 0.02 and 0.05

**thus range of p-value = 0.02 < p-value <
0.05**

Part c. At = .05, can your null hypothesis be rejected? What is your conclusion?

p-value **is less than** 0.05, We
**reject** H0.

We **conclude** that the mean annual administrator
salary in Ohio **is different from** the national mean
annual salary.

Part d. Repeat the preceding hypothesis test using the critical value approach

find t critical value for df = 24 and two tail area = 0.05 and find t critical value.

t critical value = 2.064 = 2.06

Since this is two tailed test, **t critical values are (
-2.06 , 2.06)**

Since t test statistic value = -2.141 < -2.06 , we reject H0.

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