**M5A2 Project Part 4: Inferential Statistics Report**

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# M5A2 Project Part 4: Inferential Statistics Report

**I. Introduction:**

**A.** The purpose of the survey was to identify the television viewing habits of American Adults. There was a specific focus on sports, with an emphasis on the 2019 Super Bowl, in which the New England Patriots were competing against the Los Angeles Rams. The sample group was comprised of 152 friends, co-workers, and family. The participants were people known to the surveyors who were easily reached and afforded the group the greatest opportunity of collecting relevant data. For data collection, the surveyors used one of the main types of non-probability sampling, convenience sampling. Random sampling was not the sampling method in the data collection process.

**B. We asked the following four questions:**

1. In a typical week, how many hours of TV do you watch?

2. In a typical week, how many hours of sports programs do you watch?

3. Do you watch football?

4. Who do you think will win the 2019 Super Bowl?

**II. Looking at Categorical Variable:**

**A. **For the response of our question "Do you watch football," see pie chart below.

From the pie chart above we ascertain that out of the 152 people we surveyed, 54 (35.53%) did not watch football and 98 (64.47%) did watch football.

**B. One sample proportion confidence interval:**

Outcomes in : Do You Watch Football

Success : Y

p : Proportion of successes

Method: Standard-Wald**95% confidence interval results:**

Variable | Count | Total | Sample Prop. | Std. Err. | L. Limit | U. Limit |
---|---|---|---|---|---|---|

Do You Watch Football | 98 | 152 | 0.64473684 | 0.038819017 | 0.56865297 | 0.72082072 |

**Interpretation of the confidence interval results: ** As we can see in the 95% confidence interval result box above, the variable was "Do you watch football." Our team's sample size was 152, n=152. In this particular case, 95% confidence interval means that if we were to select varying samples of this sample size of 152 (n=152), approximately 95% of these samples would still contain the true proportion p. As we can see in our results above, we were able to ascertain with 95% confidence that the true proportion of our surveyed population would answer "yes" to "Do you watch football," is between 0.569 to 0.721.

The margin of error is E=(0.721-0.645)/2 = 0.076/2 = 0.038 (See results as above).

**III. Looking at a numerical value:**

**A. **The responses to the questions "In a typical week, how many hours of sports programs do you watch" can be seen in the histogram and summary statistics box below.

### Summary statistics:

### Summary statistics:

Column | n | Mean | Variance | Std. dev. | Median | Range | Min | Max | Q1 | Q3 |
---|---|---|---|---|---|---|---|---|---|---|

Hours of Sports Watched | 152 | 3.0263158 | 14.555594 | 3.8151795 | 2 | 20 | 0 | 20 | 0 | 5 |

**B.** A 95% confidence interval for the population mean is shown below:

μ : Mean of variable**95% confidence interval results:**

Variable | Sample Mean | Std. Err. | DF | L. Limit | U. Limit |
---|---|---|---|---|---|

Hours TV Watched | 14.309211 | 0.84567429 | 151 | 12.638328 |
15.980093 |

A 95% confidence interval means that if we were to select many varying samples of size n=152, approximately 95% of these samples would result in confidence intervals that still will contain the true proportion mean. Based on data above, we interpret the results by saying we are 95% confident that our confidence interval contains the true population mean. We are 95% confident that within our population, the average person watches TV between 12.6 and 16.0 hours per week.

The T distribution was used because the population standard deviation is unknown to us.

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