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8.9: Office Temperature

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    Let’s do another example to solidify our understanding. Let’s say that the office building you work in is supposed to be kept at 74 degree Fahrenheit but is allowed to vary by 1 degree in either direction. You suspect that, as a cost saving measure, the temperature was secretly set higher. You set up a formal way to test your hypothesis.

    Step 1: State the Hypotheses

    You start by laying out the null hypothesis:

    \(H_0\): The average building temperature is not higher than claimed

    \(H_0: \mu \leq 74\)

    Next you state the alternative hypothesis. You have reason to suspect a specific direction of change, so you make a one-tailed test:

    \(H_A\): The average building temperature is higher than claimed

    \(\mathrm{H}_{\mathrm{A}}: \mu>74\)

    Step 2: Find the Critical Values

    You know that the most common level of significance is \(α\) = 0.05, so you keep that the same and know that the critical value for a one-tailed \(z\)-test is \(z*\) = 1.645. To keep track of the directionality of the test and rejection region, you draw out your distribution:

    fig 7.9.1.png
    Figure \(\PageIndex{1}\): Rejection region

    Step 3: Calculate the Test Statistic

    Now that you have everything set up, you spend one week collecting temperature data:

    Table \(\PageIndex{1}\): Temperature data for a week
    Day Temperature
    Monday 77
    Tuesday 76
    Wednesday 74
    Thursday 78
    Friday 78

    You calculate the average of these scores to be \(M\)= 76.6 degrees. You use this to calculate the test statistic, using \(μ\) = 74 (the supposed average temperature), \(σ\) = 1.00 (how much the temperature should vary), and \(n\) = 5 (how many data points you collected):

    \[z=\dfrac{76.60-74.00}{1.00 / \sqrt{5}}=\dfrac{2.60}{0.45}=5.78 \nonumber \]

    This value falls so far into the tail that it cannot even be plotted on the distribution!

    fig 7.9.2.png
    Figure \(\PageIndex{2}\): Obtained \(z\)-statistic

    Step 4: Make the Decision

    You compare your obtained \(z\)-statistic, \(z\) = 5.77, to the critical value, \(z*\) = 1.645, and find that \(z > z*\). Therefore you reject the null hypothesis, concluding:

    Based on 5 observations, the average temperature (\(M\)= 76.6 degrees) is statistically significantly higher than it is supposed to be, \(z\) = 5.77, \(p\) < .05.

    Because the result is significant, you also calculate an effect size:

    \[d=\dfrac{76.60-74.00}{1.00}=\dfrac{2.60}{1.00}=2.60 \nonumber \]

    The effect size you calculate is definitely large, meaning someone has some explaining to do!

    Contributors and Attributions

    • Foster et al. (University of Missouri-St. Louis, Rice University, & University of Houston, Downtown Campus)

    This page titled 8.9: Office Temperature is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Foster et al. (University of Missouri’s Affordable and Open Access Educational Resources Initiative) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.