Confidence Intervals and Tests of Hypotheses for Means Study Guide

Example

A study was conducted to determine the mean amount that hair grows daily. A sample of 35 young adults (from 18 to 35 years of age) was randomly selected from all patrons at a large hair salon. The amount of daily hair growth was recorded for each person in the sample. The sample mean was 0.35 mm, and the sample standard deviation was 0.05 mm. Set a 90% confidence interval on the mean daily hair growth of young adults. Interpret the interval in the context of the problem.

Solution

First, we need to determine whether the conditions for inference are satisfied. The first condition that the sample be a random one is satisfied. Although we are not told that the population distribution of daily hair growth is normally distributed, this would seem to be a reasonable assumption. If the data were available, we could construct a histogram and a boxplot to determine whether the assumption of normality is reasonable. However, because n = 35 > 30, we can appeal to the Central Limit Theorem and assume that the sampling distribution of is approximately normal.

Because the standard deviation is unknown, we use the following confidence interval:

We are given that n = 35, = 0.35 mm, and s = 0.05 mm. The t* value we need would have n –1 = 34 degrees of freedom and = = 0.025 probability in the upper tail. Looking in the t-table in Lesson 12, we have t* = 2.032. The confidence limits are 0.35 ± 2.032 or 0.35 ± 0.02. That is, we are 95% confident that the mean daily hair growth of young adults who are patrons of this salon is between 0.33 and 0.37 mm. Note: Because the population of young adult patrons of this salon was the one from which the sample was selected, it is the one to which we can draw inference. Does the inferene extend to all young adults at least in some region about the salon? It does if the patrons of the salon do not differ significantly from others in the region with respect to hair growth. Careful thought should be given before inferences are extended to a broader population than that sampled.

Hypothesis Testing for a Mean

As with proportions, there are five steps to conducting a test of hypotheses. We will consider each separately.

Step 1: Specifying the Hypotheses

Remember that, when establishing a set (H₀ and H_a) of statistical hypotheses, the research hypothesis is the alternative hypothesis. When working with means, the null hypothesis is that the population mean μ is equal to some value μ₀ The alternative may be that μ is less than, greater than, or equal to μ₀,depending on what the research hypothesis is.

Step 2: Verify Necessary Conditions for a Test and, if Satisfied, Construct the Test Statistic

The conditions for testing hypotheses about the population mean μ are the same as those for constructing a confidence interval on this parameter. If the population distribution is normal, it is enough to know that we have a random sample from the population distribution. If the population distribution is not normal, in addition to having a random sample from the population, the sample size must be large enough to assume that the sampling distribution of is approximately normal (i.e., n ≥ 30).

If the population is normal and the standard deviation is unknown or if the population distribution is not known but a large sample (n ≥ 30) has been selected, the test statistics is where μ₀ is the value of the mean under the null hypothesis.

Notice that the first test statistics has the form .

If the standard deviation of the population and hence the standard deviation of the point estimate are known, we certainly want to use the true value of the parameter instead of the standard error, its estimate. This would lead us to use a different test statistic from that presented here. The reality is that, in practice, the population is rarely, if ever, known, and we will only consider this case.

Step 3: Find the p-Value Associated with the Test Statistic

If the population distribution is normal, the test statistic has a t-distribution with (n – 1) degrees of freedom. Similarly, if the population distribution is not normal but a large sample has been selected, the test statistic has an approximate t-distribution. If the null hypothesis is not true, the test statistic does not have a t-distribution. The observed value of the test statistic t_T is likely to be unusual for a randomly selected observation from the t-distribution. The p-value continues to be the probability of observing a value as extreme as or more extreme than the test statistic, t_T, from a random selection of an observation from a t-distribution with (n – 1) degrees of freedom.

How do we measure how unusual a test statistic is? It depends on the alternative hypothesis. These are summarized in Table 17.1.

Step 4: Decide Whether or Not to Reject the Null Hypothesis

Before beginning the study, the significance level α of the test is set. If the p-value is less than the significance level, the null hypothesis is rejected; otherwise, the null is not rejected.

Step 5: State Conclusions in the Context of the Study

Statistical tests of hypotheses are conducted to determine whether or not sufficient evidence exists to reject the null hypothesis in favor of the alternative hypothesis.

Examples of Hypothesis Testing for a Mean

Below are two examples of Hypothesis Testing for a Mean using the five-steps previously discussed.

Example 1:

A machine is used to produce bolts that have a mean diameter of 1.50 cm. The diameters of the bolts are normally distributed. Sometimes, the machine produces bolts that differ from the desired 1.50 cm mean diameter. If the mean bolt diameter is significantly greater than 1.50 cm, the machine is to be reset. This is expensive because no bolts can be produced while it is being reset. A random sample of 15 bolts is selected from each day's production and tested, and their diameters are recorded. Today, the average diameter of the sampled bolts was 1.52 cm, and the sample standard deviation was 0.01 cm. Is there sufficient evidence to conclude that the machine needs to be reset?

Solution 1:

We follow the five steps of hypothesis testing.

Step 1: Specifying the Hypotheses

Here, we want strong evidence that the machine needs to be reset. This leads to the following set of hypotheses:

H₀:μ = 1.50

H_a:μ > 1.50

Notice that here we have chosen μ = 1.50 for the null hypothesis because that is the mean diameter of bolts that the machine should be producing. We want to know if the mean diameter becomes significantly greater than 1.50, so μ > 1.50 becomes the alternative hypothesis.

Step 2: Verify Necessary Conditions for a Test and, if Satisfied, Construct the Test Statistic

The population distribution of bolt diameters is normal, and the bolts have been randomly selected from the day's production. Thus, the conditions for a test have been satisfied. The test statistic is:

= = 7.75

Step 3: Find the p-Value Associated with the Test Statistic

Assuming that the population distribution is normal and the standard deviation is known, the test statistic has a t-distribution with n – 1 = 14 degrees of freedom if the null hypothesis is true. We want to find the probability that a test statistic as extreme or more extreme would be observed if the null hypothesis is not true. Because we are interested only if the sample mean is too large (see the alternative hypothesis), we find

p = P(t > t_T)

p = P(t > 7.75) = 1 – P(t ≤ 7.75) > 0.001

Step 4: Decide Whether or Not to Reject the Null Hypothesis

This p-value is so small that it will be less than any reasonable significance level. Therefore, we would reject the null hypothesis.

Step 5: State Conclusions in the Context of the Study

We reject the null hypothesis and conclude that the mean diameter of bolts produced on that day exceeds 1.50 cm. Therefore, the machine needs to be reset.

Example 2:

A researcher has read that U.S. children aged 10 to 17 years watch television an average of 3.6 hours per day. She wants to know if her region differs from this national average. She randomly selects 40 children aged 10 to 17 from the region. Because children do not watch exactly the same amount of television every day, she records the number of hours each sampled child watches television for a four-week period. By taking the average of the hours spent watching television over these 28 days, she has a good measure of the amount of time each child spends watching television daily. The sample mean number of hours these 40 children watched television daily is 2.8 hours, and the sample standard deviation is 2.2 hours. Is there sufficient evidence to conclude that the average number of hours children spend watching television in the researcher's region is different from that reported for the nation?

Solution 2:

Again, we follow the five steps of hypothesis testing.

Step 1: Specifying the Hypotheses

Here, we want to know whether the average for the region differs from that for the nation. The direction of that difference is not specified, so we have a two-sided alternative; that is, we want to test the following set of hypotheses:

H₀:μ = 3.6

H_a:μ ≠ 3.6

Step 2: Verify Necessary Conditions for a Test and, if Satisfied, Construct the Test Statistic

A random selection of children was taken from all children in the region (the population of interest). The population distribution of the number of hours children spend watching television daily is very unlikely to be normal. However, because the sample size is 40 (> 30), we know that has at least an approximate normal distribution by the Central Limit Theorem. Therefore, the two conditions for inference are satisfied. The test statistic is

= = 1.15.

Step 3: Find the p-Value Associated with the Test Statistic

The test statistic has an approximate t-distribution with n – 1 = 39 degrees of freedom if the null hypothesis is true. We want to find the probability that a test statistic as extreme or more extreme would be observed if the null hypothesis is not true. Because we are interested if the sample mean gets "too far" from the hypothesized population mean, we f i n d p =P(|t| > |t_T|) = p =P(|t_T| > 1.15) > 2 × 0.1 = 0.2 using the table in Lesson 12. Notice that the smallest t-value in the line for 39 degrees of freedom in the t-table is 1.304, corresponding to a right-tail probability of 0.01. Because 1.15 is smaller than 1.304, more area is to the right of 1.15 than to the right of 1.304. Thus, P(t > 1.15) > 0.1.Because the alternative hypothesis is two sided, this right-tail probability must be doubled to account for the values in the left-tail that would have also been at least as extreme as what we observed if the null hypothesis is true.

Step 4: Decide Whether or Not to Reject the Null Hypothesis

The p-value being greater than 0.2 indicates that what we observed would not be at all unusual if the null hypothesis is true (p > α).Therefore, we would not reject the null hypothesis.

Step 5: State Conclusions in the Context of the Study

We fail to reject the null hypothesis that the mean number of hours children in the researcher's region watch television is the same as that for children in the nation.

Confidence Intervals and Tests of Hypotheses for Means In Short

The sample mean is the best estimate of the population mean, but the two are rarely equal. Thus, we use confidence intervals to establish an interval of values that will capture the true population mean with a specified level of confidence. The form of these intervals continues to be estimate ± multiplier× standard error as it was for proportions. We may also want to test hypotheses concerning values of the population mean. The test statistic for means is , where μ₀ is the hypothesized value of the population mean.

Find practice problems and solutions for these concepts at Confidence Intervals and Tests of Hypotheses for Means Practice Questions.