# In a measure of well-being, th

In a measure of well-being, the state in which I live had a meanwell-being score of 65.5 with a standard deviation of 1.6. I wonderif the people within my neighborhood would be significantlydifferent from the state statistics. So I calculate the well-beingscores for 16 within my neighborhood to run a statisticalcomparison. The data are presented below. What is the dependentvariable and independent variable in this statistical test?

Subject |
Well-being Score |

1 |
68 |

2 |
78 |

3 |
64 |

4 |
71 |

5 |
67 |

6 |
66 |

7 |
68 |

8 |
57 |

9 |
65 |

10 |
49 |

11 |
68 |

12 |
71 |

13 |
69 |

14 |
64 |

15 |
68 |

16 |
66 |

Group of answer choices

DV-My neighborhood; IV-Level ofwell-being

DV-Well-being; IV-state

DV-Location (state or myneighborhood); IV-Level of well-being

DV-Level of well-being; IV-Location(state or my neighborhood)

Question 2 1 pts

What would be the null hypothesis formy statistical test? Note that I am interested in examining anydifference between my sample and the population statistics and I amnot predicting or hypothesizing any particular direction of thedifference.

Group of answer choices

Level of well-being in my neighborhoodis not significantly different from the general level of well-beingin my state.

Level of well-being in my neighborhoodis significantly different from the general level of well-being inmy state.

Level of well-being in my neighborhoodis significantly higher from the general level of well-being in mystate.

Level of well-being in my neighborhoodis not significantly higher from the general level of well-being inmy state.

Question 3 1 pts

In order to compare the sample Icollected (the well-being scores from my neighborhood) to the statestatistics, calculate the mean of the sample. (Round it to 1decimal place.)

Question 4 1 pts

Because a sample should be compared toa population of samples, we need to turn the original well-beingscores from the state into a sampling distribution, which iscomposed of all the possible samples drawn from the whole statepopulation of well-being scores. In order to position my sample onthat sampling distribution, we need to know the standard deviationof the sampling distribution, which is also commonly calledstandard error (SE). Calculate SE from the state well-beingstatistics provided in the research scenario: mean (μ) = 65.5, SD(σ) = 1.6, along with the sample size N = 16. (Round it to 1decimal place)

Question 5 1 pts

Now we need to locate our sample meanon the sampling distribution and see if it falls into an extreme(significance) area. If so, we would be able to reject the nullhypothesis and say that the well-being scores in my neighborhoodare indeed significantly different from the state well-beingscores. To figure out the position of our sample mean on thesampling distribution, calculate the Z statistic for our sample.(Report to two decimal places.)

Question 6 1 pts

In the previous question, wecalculated the Z statistic, which indicates where our sample meanis located in the sampling distribution. Now we need to find thecritical Z value from the Z table so that we can see if the Zstatistic (calculated from the sample mean) is more extreme thanthe critical Z value (the cut-off point) for the significance area.But before identifying the critical Z value, we need to firstdetermine whether we will have the significance area in one tail ortwo tails, based on the hypotheses? In other words, isthe hypothesis test two-tailed or one-tailed?

Group of answer choices

Two-tailed

One-tailed

Either one-tailed or two-failed wouldwork

We don’t have sufficient informationto make the decision.

Question 7 1 pts

Use the Z table to identify thecritical Z value(s) for the type of hypothesis test selected, basedon α = .05.

Group of answer choices

+1.96 and -1.96

+1.645 and -1.645

+1.96

+1.645

Question 8 1 pts

Now we determine whether our Zstatistic (calculated from the sample mean) would fall into thesignificance area of the comparison sampling distribution, bycomparing the Z statistic to the critical Z value (cut-off). Basedon the Z statistic and critical Z value obtained in previousquestions, what is the result of the hypothesis test?

Group of answer choices

We fail to reject the null hypothesisbecause the calculated Z is less extreme than the critical Zvalue.

We fail to reject the null hypothesisbecause the calculated Z is more extreme than the critical Zvalue.

We reject the null hypothesis becausethe calculated Z is less extreme than the critical Z value.

We reject the null hypothesis becausethe calculated Z is more extreme than the critical Z value.

Question 9 1 pts

The decision of rejecting the nullhypothesis or failing to reject the null hypothesis does notdirectly answer the research question so it is always necessary toreiterate the analysis result by directly answering the researchquestion. Which of the following answers is consistent with thehypothesis test result?

Group of answer choices

The well-being scores in myneighborhood are not significantly different from the well-beingscores in the state.

There is no significant differencebetween neighborhood and well-being scores.

My neighborhood is significantlydifferent from the state in terms of well-being scores.

The well-being scores in myneighborhood are significantly higher than the general well-beingscores in the state.

Question 10 1 pts

The result regarding significance onlytells us whether the difference between the two populations beingcompared is statistically significant, but it does not tell us howbig the potential difference is. Calculate the standardized effectsize for this test, using the information provided in the researchscenario and the answers from previous questions. (Round the answerto 2 decimal places.)

Answer:

DV-Level of well-being; IV-Location (state or myneighborhood)

Question 2 1 pts

Level of well-being in my neighborhood is not significantlydifferent from the general level of well-being in my state.

Question 3 1 pts

From the data, mean of the sample is,

Mean ()= (68 + 78 + 64 + 71 + 67 + 66 + 68 + 57 + 65 + 49 + 68 + 71 + 69 +64 + 68 + 66) / 16 = 66.2

Question 4 1 pts

SE = = = 0.4

Question 5 1 pts

Z = (– ) / SE = (66.2 – 65.5) / 0.4 = 1.75

Question 6 1 pts

Since we are interested in examining any difference between mysample and the population statistics not predicting orhypothesizing any particular direction of the difference, we useTwo-tailed test.

Question 7 1 pts

For α = .05 and two-tailed test, the Z value is,

+1.96 and -1.96

Question 8 1 pts

Since the calculated Z statistic (1.75) is less than +1.96,

We fail to reject the null hypothesis because the calculated Zis less extreme than the critical Z value.

Question 9 1 pts

Since, we fail to reject the null hypothesis,

The well-being scores in my neighborhood are not significantlydifferent from the well-being scores in the state.

Question 10 1 pts

Effect Size = (– ) / = (66.2 – 65.5) / 1.6 = 0.44