## Question 857:

1## Answer:

No answer provided yet.We will use a Chi-Square test of independence to determine if there is a relationship between packaging size and economic status. The Null Hypothesis for this test is that there is no relationship between the two variables (they are independent). We will reject the Null Hypothesis if we obtain a p-value of less than .05.The test statistic, Chi-Square, is found by subtracting the expected value from the observed value, squaring this difference, then dividing by the expected value. You add this value up for all the cells in the table.

First we compute the column totals and row totals.

Lower | Middle | Upper | Row Total | |

Small | 24 | 22 | 18 | 64 |

Medium | 23 | 28 | 19 | 70 |

Large | 18 | 27 | 29 | 74 |

Jumbo | 16 | 21 | 33 | 70 |

81 | 98 | 99 | 278 |

Expected Values

Expected values are found by multiplying the row total by the column total and dividing by the total for the table.

Lower | Middle | Upper | |

Small | 18.647 | 22.56 | 22.79 |

Medium | 20.39 | 24.67 | 24.93 |

Large | 21.56 | 26.08 | 26.35 |

Jumbo | 20.39 | 24.67 | 24.92 |

Now we subtract the expected counts from the actual counts from the first table, square it and divide by the expected.

Lower | Middle | Upper | |

Small | 1.53 | .013 | 1.007 |

Medium | .3325 | .447 | 1.41 |

Large | .588 | .032 | .2659 |

Jumbo | .947 | .547 | 2.61 |

Finally we add up all the values in the 3rd table to get the Chi-Square test statistic of 9.74. We evaluate its significance using the degrees of freedom found as the (# of rows-1)*(# of cols -1) = 2*3 = 6. Looking up the significance of the chi-square value using a table or the excel function = CHIDIST(9.74,6) we get the p-value of .1359.

Since the p-value is above the alpha cutoff of .05 we FAIL to reject the Null and conclude there is not enough evidence to conclude the variables are related. In other words, there is not a significant relationship between packaging preference and economic status.