Question

The time married men with children spend on child care averages hours per week. You belong to a professional group on family practices that would like to do its own study to determine if the time married men in your area spend on child care per week differs from the reported mean of 6.8 hours per week. A sample of 40 married couples will be used with the data collected showing the hours per week the husband spends on child care. The sample data are contained in the table below.

10.5 | 4.3 | 5.2 | 3.1 |

4.5 | 5.5 | 7.2 | 11.3 |

8.8 | 9.2 | 2.3 | 10.9 |

3.9 | 9.2 | 4.1 | 4.7 |

9.9 | 9.5 | 11.8 | 5.2 |

8.1 | 11.1 | 8.4 | 4.3 |

0.9 | 10.3 | 2.5 | 8.0 |

9.3 | 1.9 | 5.0 | 10.1 |

5.2 | 10.9 | 5.4 | 4.4 |

11.4 | 10.8 | 10.7 | 9.8 |

**a.** What are the hypotheses if your group would like to determine if the population mean number of hours married men are spending in child care differs from the mean reported in your area?

- Select your answer -<>≤≥=≠Item 1 |

- Select your answer -<>≤≥=≠Item 2 |

**b.** What is the sample mean (to 1 decimal)?

Calculate the value of the test statistic (to 2 decimals).

What is the -value?

- Select your answer -lower than .01between .01 and .025between .025 and .05between .05 and .10between .10 and .20greater than .20Item 5

**c.** Select your own level of significance. What is your conclusion?

- Select your answer -Do not rejectRejectItem 6 the null. We - Select your answer -don't havehaveItem 7 enough evidence to disprove the reported time.

Step 1

(a) Null and Alternative Hypotheses:

The following null and alternative hypotheses need to be tested:

Ho: μ = 6.8

Ha: μ ≠ 6.8

This corresponds to a Two tail test, for which a z-test for one mean should be used.

Step 2

b) First we have to find sample mean and sample standard deviation.

We are creating a table for above calculation:

Step 3

Sample mean:

Formula:

So sample mean = 7.24

Sample standard deviatio...

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