Question

A researcher is interested in determining the average SAT score
to be expected of a group of test takers. They collect data from 12
randomly selected high school seniors and construct a 98% conﬁdence
interval. Assume the SAT scores are normally distributed. [If you
do not have a calculator: Px = 21170, Px2 = 38047900] 1700 1940
1510 2000 1430 1870 1990 1650 1820 1670 2210 1380

(a) To create the desired conﬁdence interval, should we be using a
T or Z distribution? Why? (b) Find the sample mean, sample standard
deviation and the degrees of freedom if necessary. (c) Give the
margin of error and construct the conﬁdence interval.

Answer #1

a. As sample size is n<30 so we will use t distribution to find CI

b.

Create the following table.

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

1700 | -64.1667 | 4117.36538889 |

1940 | 175.8333 | 30917.34938889 |

1510 | -254.1667 | 64600.71138889 |

2000 | 235.8333 | 55617.34538889 |

1430 | -334.1667 | 111667.38338889 |

1870 | 105.8333 | 11200.68738889 |

1990 | 225.8333 | 51000.67938889 |

1650 | -114.1667 | 13034.03538889 |

1820 | 55.8333 | 3117.35738889 |

1670 | -94.1667 | 8867.36738889 |

2210 | 445.8333 | 198767.33138889 |

1380 | -384.1667 | 147584.05338889 |

Find the sum of numbers in the last column to get.

So

df is 12-1=11

c. t value for 11 df for 98% CI is TINV(0.02,11)=2.718

So Margin of Error is

Hence CI is

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 4 minutes ago

asked 4 minutes ago

asked 5 minutes ago

asked 14 minutes ago

asked 46 minutes ago

asked 51 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago