Question

The state test scores for 12 randomly selected high school seniors are shown on the right. Complete parts (a) through (c) below. Assume the population is normally distributed. 1420 1227 989 691 726 833 722 747 547 630 1442 948 (a) Find the sample mean. x overbarequals nothing (Round to one decimal place as needed.) (b) Find the sample standard deviation. sequals nothing (Round to one decimal place as needed.) (c) Construct a 90% confidence interval for the population mean mu. A 90% confidence interval for the population mean is (

Answer #1

n = 12

Data : 1420, 1227, 989, 691, 726, 833, 722, 747, 547, 630, 1442, 948

(a) Mean = sum of terms / no of terms = 10922 / 12 = 910.1667

**Mean = 910.2**

(b) standard deviation = s

data | data-mean | (data - mean)2 |

1420 | 509.8333 | 259929.99378889 |

1227 | 316.8333 | 100383.33998889 |

989 | 78.8333 | 6214.68918889 |

691 | -219.1667 | 48034.04238889 |

726 | -184.1667 | 33917.37338889 |

833 | -77.1667 | 5954.69958889 |

722 | -188.1667 | 35406.70698889 |

747 | -163.1667 | 26623.37198889 |

547 | -363.1667 | 131890.05198889 |

630 | -280.1667 | 78493.37978889 |

1442 | 531.8333 | 282846.65898889 |

948 | 37.8333 | 1431.35858889 |

**Standard deviation = 303. 2**

(c) Confidence interval

t value for 90% of confidence interval at 11 df is TINV(0.10,11) = 1.796

CI = mean +/- E = (753.013 , 1067.387)

**CI = (753.0 , 1067.4)**

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