# 7.5 - Confidence Intervals for Means

Previously we considered confidence intervals for 1-proportion and our multiplier in our interval used a *z*-value. But what if our variable of interest is a quantitative variable (e.g. GPA, Age, Height) and we want to estimate the population mean? In such a situation proportion confidence intervals are not appropriate since our interest is in a **mean** amount and not a proportion.

We apply similar techniques when constructing a confidence interval for a mean, but now we are interested in estimating the population mean (\(\mu\)) by using the sample statistic (\(\overline{x}\)) and the multiplier is a *t* value. At the end of Lesson 6 you were introduced to this *t* distribution. Similar to the *z* values that you used as the multiplier for constructing confidence intervals for population proportions, here you will use *t* values as the multipliers. Because *t* values vary depending on the number of degrees of freedom (df), you will need to use either the *t* table or statistical software to look up the appropriate *t* value for each confidence interval that you construct. Using either method, the degrees of freedom will be based on the sample size, *n*. Since we are working with one sample here, \(df=n-1\).

### Finding the *t** Multiplier

Reading the *t *table is slightly more complicated than reading the *z *table because for each different degree of freedom there is a different distribution. In order to locate the correct multipler on the *t *table you will need two pieces of information: (1) the degrees of freedom and (2) the confidence level. The columns of the *t* table are for different confidence levels (80%, 90%, 95%, 98%, 99%, 99.8%). The rows of the *t* table are for different degrees of freedom. The multiplier is at the intersection of the two.

### Examples

**Cups of Coffee**

A research team wants to estimate the number of cups of coffee the average Penn State student consumes each week with 95% confidence. They take a random sample of 20 students and ask how many cups of coffee they drink each week.

**Average Height**

Sports analysts are studying the heights of college quarterbacks. They take a random sample of 55 college quarterbacks and measure the height of each. They want to construct a 98% confidence interval.

Our confidence level is 98%. \(df=55-1=54\)

Our *t* table does not provide us with multipliers for 54 degrees of freedom. To be more conservative, we will use 50 degrees of freedom because that will give us the larger multiplier.

Using the *t* table, our multiplier will be 2.403

You can also use statistical software to look up *t** multipliers.

### Finding *t* *Multipliers with Minitab Express and Minitab

To find the *t*-multipliers in Minitab Express:

- Probability > Probability Distribution > Display Probability
- Select
*t*distribution and enter your degrees of freedom (below we used df=15) - Select "A Specified Probability" and "Equal Tails"
- The probability is equal to alpha (i.e., \(1 - confidence \;level\); below we used a 98% confidence interval, so 0.02 in the tails combined)
- Click Ok, the values at the bottom of the graph (seen below) are your multipliers.

**Video Review- No sound**

To find the *t*-multipliers in Minitab:

- Graph > Probability Distirbution Plot > View Probability
- Change "Distribution" to
*t*and enter your degrees of freedom - Click the "Shaded Area" tab and select "Both Tails," the proportion in both tails will be equal to alpha (i.e., \(1 - confidence \;level\))
- For example, for a 95% confidence interval, the proportion in both tails would equal \(1-.95=.05\). You would enter .05
- Click Ok, the values at the bottom of the graph are your multipliers.

### Constructing a Confidence Interval for \(\mu\)

Let’s review some of symbols and equations that we learned in previous lessons:

Sample size | \(n\) |

Population mean | \(\mu=\frac{\sum X}{N}\) |

Sample mean | \(\overline{x}= \frac{\sum x}{n}\) |

Standard error of \(\mu\) | \(SE(\overline{x})=\frac{s}{\sqrt{n}}\) |

Multiplier | \(t^{*} \) |

Degrees of freedom (one group) | \(df=n-1\) |

Recall from earlier this lesson, the general form for a confidence interval is \(point\;estimate\pm(multiplier)(standard\;error)\)

For a population mean, the point estimate is \(\overline{x}\), the standard error is \(SE(\overline{x})\) and the multiplier is \(t^{*} \). When we put these together, the formula for a confidence interval for a population mean is

**Confidence Interval for a Population Mean**

\(\overline{x} \pm t^{*} \frac{s}{\sqrt{n}}\)

### Example: Mean Pitcher's Age

In a sample of 30 current MLB pitchers, the mean age was 28 years with a standard deviation of 4.4 years. Construct a 95% confidence interval to estimate the mean age of all current MLB pitchers.

### Example: Sleep Deprivation

In a class survey, students are asked how many hours they sleep per night. In the sample of 22 students, the mean was 5.77 hours with a standard deviation of 1.572 hours. Let’s construct a 95% confidence interval for the mean number of hours slept per night in the population from which this sample was drawn.

This is what we know: \(n=22\), \(\overline{x}=5.77\), and \(s=1.572\).

In order to compute the confidence interval for \(\mu\) we will need the *t* multiplier and the standard error (\( \frac{s}{\sqrt{n}}\)).

\(df=n-1=22-1=21\)

For a 95% confidence interval with 21 degrees of freedom, \(t^{*}=2.080\)

\(SE(\overline{x})=\frac{s}{\sqrt{n}}=\frac{1.572}{\sqrt{22}}=0.335\)

Thus, our confidence interval for \(\mu\) is: \(5.77\pm 2.080(0.335)=5.77\pm0.697=[5.073,\;6.467]\)

We are 95% confident that the population mean is between 5.073 and 6.467 hours.

**What if we wanted to be more conservative and construct a 99% confidence interval?**

The only thing that would change is our multiplier. Now, \(t^{*}=2.831\).

\(5.77\pm 2.831(0.335)=5.77\pm0.948=[4.822,\;6.718]\)

We are 99% confident that the population mean is between 4.822 and 6.718 hours.

### Example:Milk Production

A study of 66,831 dairy cows found that the mean milk yield was 12.5 kg per milking with a standard deviation of 4.3 kg per milking (data from Berry, et al., 2013). Construct a 95% confidence interval for the average milk yield in the population.

\(SE(\overline{x})=\frac{s}{\sqrt{n}}=\frac{4.3}{\sqrt{66831}}=0.0166\)

The standard error is small because the sample size is very large.

\(df=66831-1=66830\)

As degrees of freedom approach infinity, the *t* distribution approaches the *z* distribution. Our *t* table only goes to \(df=100\), so we can use the last line where \(df=infinity\).

\(t^{*}=1.96\)

95% C.I.: \(12.5\pm1.96(0.017)=12.5\pm0.033=[12.467,\;12.533]\)

We are 95% confident that the mean milk yield in the population is between 12.467 and 12.533 kg per milking.

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Berry, D. P., Coyne, J., Boughlan, B., Burke, M., McCarthy, J., Enright, B., Cromie, A. R., McParland, S. (2013). Genetics of milking characteristics in dairy cows. Animal, 7(11), 1750-1758.