Question

Given two dependent random samples with the following results:

Population 1 | 19 | 26 | 26 | 30 | 48 | 33 | 31 |
---|---|---|---|---|---|---|---|

Population 2 | 24 | 38 | 40 | 38 | 39 | 41 | 29 |

Use this data to find the 98% confidence interval for the true difference between the population means.

Let d=(Population 1 entry)−(Population 2 entry)d=(Population 1 entry)−(Population 2 entry). Assume that both populations are normally distributed.

Step 1 of 4 : Find the mean of the paired differences, x‾d. Round your answer to one decimal place.

Step 2 of 4: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Step 3 of 4: Find the standard deviation of the paired differences to be used in constructing the confidence interval. Round your answer to one decimal place.

Step 4 of 4: Construct the 98% confidence interval. Round your answers to one decimal place.

Answer #1

Population 1 |
Population 2 |
Difference |

19 | 24 |
-5 |

26 | 38 |
-12 |

26 | 40 |
-14 |

30 | 38 |
-8 |

48 | 39 |
9 |

33 | 41 |
-8 |

31 | 29 |
2 |

1)

∑d = -36

∑d² = 578

n = 7

Mean , x̅d = Ʃd/n = -36/7 = -5.1429 = **-5.1**

2)

At α = 0.02 and df = n-1 = 6, two tailed critical value, t-crit
= T.INV.2T(0.02, 6) = **3.143**

3)

Standard deviation, sd = √[(Ʃd² - (Ʃd)²/n)/(n-1)] =
√[(578-(-36)²/7)/(7-1)] = 8.0917 = **8.1**

4)

98% Confidence interval :

At α = 0.02 and df = n-1 = 6, two tailed critical value, t-crit = T.INV.2T(0.02, 6) = 3.143

Lower Bound = x̅d - t-crit*sd/√n = -5.1429 - 3.143 * 8.0917/√7 = -14.8

Upper Bound = x̅d + t-crit*sd/√n = -5.1429 + 3.143 * 8.0917/√7 = 4.5

**-14.8 < µd < 4.5**

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