During the winter months, the temperatures at the Starneses’ Colorado cabin can stay well below freezing (32°F or 0°C) for weeks at a time. To prevent the pipes from freezing, Mrs. Starnes sets the thermostat at 50°F. The manufacturer claims that the thermostat allows variation in home temperature of s = 3°F. Mrs. Starnes suspects that the manufacturer is overstating the consistency of the thermostat.
The correct answer and explanation is:
Answer:
Mrs. Starnes is testing the manufacturer’s claim that the thermostat maintains the home temperature with a standard deviation of s=3∘Fs = 3^\circ F. To assess this, she collects data and calculates the sample standard deviation (ss). If the sample standard deviation is significantly greater than 3°F, it suggests the manufacturer may have overstated the thermostat’s consistency. The correct hypothesis test to use is a chi-square test for variance:
Hypothesis Setup
- Null Hypothesis (H0H_0): σ2=9\sigma^2 = 9 (the manufacturer’s claim: s=3∘Fs = 3^\circ F, so 32=93^2 = 9)
- Alternative Hypothesis (HaH_a): σ2>9\sigma^2 > 9 (Mrs. Starnes believes the variation is greater)
The test statistic for variance follows a chi-square distribution: χ2=(n−1)s2σ2\chi^2 = \frac{(n – 1) s^2}{\sigma^2}
where:
- nn is the sample size,
- s2s^2 is the sample variance,
- σ2=9\sigma^2 = 9 (claimed variance).
If the computed chi-square value is significantly large compared to the critical value from a chi-square table, we reject H0H_0, concluding that the thermostat’s variability is indeed greater than claimed.
Explanation :
In cold climates, maintaining a stable home temperature is crucial to prevent pipes from freezing. Mrs. Starnes relies on her thermostat, which the manufacturer claims holds a standard deviation of 3°F. However, she suspects that temperature fluctuations are greater.
To investigate, she records home temperatures over several weeks and calculates the sample standard deviation (ss). Since standard deviation measures spread, a higher-than-expected ss suggests the thermostat is inconsistent.
A chi-square test for variance is appropriate since variance follows a chi-square distribution when sampling from a normal population. The test compares the observed sample variance s2s^2 against the claimed variance of 9°F².
Mathematically, the chi-square statistic is: χ2=(n−1)s2σ2\chi^2 = \frac{(n – 1) s^2}{\sigma^2}
where nn is the sample size, s2s^2 is the computed variance, and σ2=9\sigma^2 = 9. The result is compared against a chi-square distribution with n−1n-1 degrees of freedom.
If the test statistic exceeds the critical value at a chosen significance level (e.g., α=0.05\alpha = 0.05), Mrs. Starnes has statistical evidence that the thermostat fluctuates more than claimed.
This test helps consumers verify manufacturer claims, ensuring reliability. If the thermostat is inconsistent, it could lead to unexpected temperature drops, risking frozen pipes. If Mrs. Starnes finds strong evidence, she might consider replacing the thermostat or taking additional precautions to protect her home during freezing weather.
Now, I will generate an image visualizing this statistical test.
Here is the generated illustration of Mrs. Starnes’ cozy log cabin in a snowy landscape, featuring the thermostat set at 50°F and a statistical overlay related to the chi-square test. Let me know if you need any modifications or additional details!
