Inference for a Population Mean (HT for 1 Mean, sigma Known)

Now we get to the good stuff! We will need to know how to label the null and alternative hypothesis, calculate the test statistic, and then reach our conclusion using the critical value method or the p-value method.

The Test Statistic for Testing 1 Mean, σ Known:

[latex]z = \displaystyle \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/latex]

What the different symbols mean:

[latex]n[/latex] is the sample size (number of people, items, etc… in the study)

[latex]\mu[/latex] is the population mean

[latex]\bar{x}[/latex] is the sample mean (also known as average)

[latex]\sigma[/latex] is the population standard deviation (sometimes we use this as an assumption)

[latex]\alpha[/latex] is the significance level, usually given within the problem, or if not given, we assume it to be 5% or 0.05

Assumptions when conducting a Test for 1 Mean, σ Known:

• We have a simple random sample
• We have a normal distribution OR [latex]n\ge 30[/latex]

Steps to conduct a Test for 1 Mean, σ Known:

• Identify all the symbols listed above (all the stuff that will go into the formulas). This includes [latex]n[/latex], [latex]\mu[/latex], [latex]\bar{x}[/latex], [latex]\sigma[/latex], and [latex]\alpha[/latex]
• Identify the null and alternative hypotheses
• Calculate the test statistic, [latex]z = \displaystyle \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/latex]
• Find the critical value(s) OR the p-value OR both
• Apply the Decision Rule
• Write up a conclusion for the test

Example 1: SNAP Benefits^[1]

According to a report from the US Department of Agriculture, the maximum monthly SNAP allowance during 2020 for one person (who could not earn income) was $194 in the U.S. An simple random sample of [latex]n = 31[/latex] U.S. grocery stores found the average cost of the essentials bundle was [latex]\bar{x} = \$199.45[/latex], with a known standard deviation of [latex]\sigma = \$25.98[/latex]. Is there convincing statistical evidence that the maximum SNAP individual monthly allowance ($194) doesn’t cover the true mean price of the essentials-only grocery bundle in the U.S.?

Solution

Since we are being asked for convincing statistical evidence, a hypothesis test should be conducted. In this case, we are dealing with averages or means from one sample or group (the grocery stores), so we will conduct a Test for 1 Mean, [latex]\sigma[/latex] Known.

• [latex]n = 31[/latex]
• [latex]\bar{x} = \$199.45[/latex]
• [latex]\sigma = \$25.98[/latex]
• [latex]\alpha = 0.05[/latex] (we were not told a specific value in the problem, so we are assuming it is 5%)
• Null and Alternative Hypothesis: Since the US Department of Agriculture report states that the allowance for one person should be $194, and the question is whether or not the maximum monthly allowance will cover the essentials-only grocery bundle, the claim that goes along with the alternative hypothesis is that [latex]\mu[/latex] is greater than $194. In our example here, the idea that “nothing is different” would be equivalent to saying that [latex]\mu[/latex] is the same as (equal to) $194.
  • [latex]H_{0}: \mu = \$194[/latex]
  • [latex]H_{A}: \mu > \$194[/latex]
• [latex]\mu = \$194[/latex] (from the null hypothesis)
• Test Statistic
  • [latex]z = \displaystyle \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\ = \displaystyle \frac{199.45 - 194}{\frac{25.98}{\sqrt{31}}} = 1.17[/latex] (remember we round z-scores to 2 places)
• P-Value: The p-value is found by looking up the test statistic calculated (in this case [latex]z = 1.17[/latex]) in the normal distribution table. We find that this corresponds to a value of [latex]0.8790[/latex]. Since this is a “greater than” test, we subtract from one, and get [latex]p-value = 1 - 0.8790 = 0.121[/latex].
• Applying the Decision Rule: We now compare this to our significance level, which is 0.05. If the p-value is smaller or equal to the alpha level, we have enough evidence for our claim, otherwise we do not. Here, [latex]p-value = 0.121[/latex], which is definitely greater than [latex]\alpha = 0.05[/latex], so we do not have enough evidence for the claim…but what does this mean?
• Conclusion: Because our p-value of [latex]0.121[/latex] is greater than our [latex]\alpha[/latex] level of [latex]0.05[/latex], we fail to reject [latex]H_{0}[/latex]. We do not have convincing evidence that the true mean price of an essentials grocery bundle in the U.S. is greater than the allowance SNAP provides.
• Follow-up commentary: But why? The sample mean came out to be larger than the allowance. It is possible to get a mean (or average) that is higher or lower than the hypothesized value (null) with some likelihood. Different neighborhoods, different stores, etc…

Example 2: Wait Times Calling the Bank

I don’t know about you, but I really hate being on hold for a long time when making a phone call. I would hope that when I call my bank, my wait time would be something reasonable, hopefully less than 5 minutes. A recent informal survey of [latex]n = 40[/latex] local and national banks revealed a mean wait time of [latex]\bar{x} = 4.2[/latex] minutes, with a known standard deviation of [latex]\sigma = 3.1[/latex] minutes. Is there convincing statistical evidence that the bank phone wait times are under 5 minutes?

Solution

Since we are being asked for convincing statistical evidence, a hypothesis test should be conducted. In this case, we are dealing with averages or means from one sample or group (the banks), so we will conduct a Test for 1 Mean, [latex]\sigma[/latex] Known.

• [latex]n = 40[/latex]
• [latex]\bar{x} = 4.2[/latex] minutes
• [latex]\sigma = 3.1[/latex] minutes
• [latex]\alpha = 0.05[/latex] (we were not told a specific value in the problem, so we are assuming it is 5%)
• Null and Alternative Hypothesis: Since my notion is that wait times should be under 5 minutes, the claim that goes along with the alternative hypothesis is that [latex]\mu[/latex] is less than 5. In our example here, the idea that “nothing is different” would be equivalent to saying that [latex]\mu[/latex] is the same as (equal to) 5.
  • [latex]H_{0}: \mu = 5[/latex]
  • [latex]H_{A}: \mu < 5[/latex]
• [latex]\mu = 5[/latex] (from the null hypothesis)
• Test Statistic
  • [latex]z = \displaystyle \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\ = \displaystyle \frac{4.2 - 5}{\frac{3.1}{\sqrt{40}}} = -1.63[/latex] (remember we round z-scores to 2 places)
• P-Value: The p-value is found by looking up the test statistic calculated (in this case [latex]z = -1.63[/latex]) in the normal distribution table. We find that this corresponds to a value of [latex]0.0516[/latex]. Since this is a “less than” test, we keep the value from the table, and get [latex]p-value = 0.0516[/latex].
• Applying the Decision Rule: We now compare this to our significance level, which is [latex]\alpha = 0.05[/latex]. If the p-value is smaller or equal to the alpha level, we have enough evidence for our claim, otherwise we do not. Here, [latex]p-value = 0.0516[/latex], which is close, but still greater than [latex]\alpha = 0.05[/latex], so we do not have enough evidence for the claim…but what does this mean?
• Conclusion: Because our p-value of [latex]0.0516[/latex] is greater than our [latex]\alpha[/latex] level of [latex]0.05[/latex], we fail to reject [latex]H_{0}[/latex]. We do not have convincing evidence that the bank phone wait times are under 5 minutes.
• Follow-up commentary: But why? The sample mean came out to be less than 5. It is possible to get a mean (or average) that is higher or lower than the hypothesized value (null) with some likelihood. Factors such as sample size, variation, and where the sample comes from, can have an effect on the outcomes.

Example 3: Average Commute Time to Work in the Imperial County

One of the possible benefits of living in the Imperial County is proximity to work. In many big cities and larger areas, commute times can hover around 25 minutes. According to recent data from the US Census Bureau, the average commute time in Imperial County for [latex]n = 5,000[/latex] workers aged 16 or older was [latex]\bar{x} = 22[/latex] minutes, and we will assume a standard deviation of [latex]\sigma = 4[/latex] minutes. Is there convincing statistical evidence that the Imperial County Commute time is under 25 minutes?

Solution

Since we are being asked for convincing statistical evidence, a hypothesis test should be conducted. In this case, we are dealing with averages or means from one sample or group (workers over the age of 16), so we will conduct a Test for 1 Mean, [latex]\sigma[/latex] Known.

• [latex]n = 5000[/latex]
• [latex]\bar{x} = 22[/latex] minutes
• [latex]\sigma = 4[/latex] minutes
• [latex]\alpha = 0.05[/latex] (we were not told a specific value in the problem, so we are assuming it is 5%)
• Null and Alternative Hypothesis: Since we are being asked if there is statistical evidence that commute times are under 25 minutes, the claim that goes along with the alternative hypothesis is that [latex]\mu[/latex] is less than 25. In our example here, the idea that “nothing is different” would be equivalent to saying that [latex]\mu[/latex] is the same as (equal to) 25.
  • [latex]H_{0}: \mu = 25[/latex]
  • [latex]H_{A}: \mu < 25[/latex]
• [latex]\mu = 25[/latex] (from the null hypothesis)
• Test Statistic
  • [latex]z = \displaystyle \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\ = \displaystyle \frac{22 - 25}{\frac{4}{\sqrt{85000}}} = -218.66[/latex] (remember we round z-scores to 2 places)
• P-Value: The p-value is found by looking up the test statistic calculated (in this case [latex]z = -218.66[/latex]) in the normal distribution table. We find that this corresponds to a value of [latex]0.0001[/latex]. Since this is a “less than” test, we keep the value from the table, and get [latex]p-value = 0.0001[/latex].
• Applying the Decision Rule: We now compare this to our significance level, which is [latex]\alpha =  0.05[/latex]. If the p-value is smaller or equal to the alpha level, we have enough evidence for our claim, otherwise we do not. Here, [latex]p-value = 0.0001[/latex], which is definitely smaller than [latex]\alpha = 0.05[/latex], so we have enough evidence for the claim…but what does this mean?
• Conclusion: Because our p-value of [latex]0.0001[/latex] is less than our [latex]\alpha[/latex] level of [latex]0.05[/latex], we reject [latex]H_{0}[/latex]. We have convincing evidence that the Imperial County Commute time is under 25 minutes.