A recent claim stated the mean temperature setting at a person’s home is 78.3 degrees F. A simple random sample of 40 homes has a mean of 77.5 degrees F. Assume the population standard deviation is 1.9 degrees F and a confidence level of 0.05. Find the P-value needed to test the claim that the mean temperature setting at a person’s home is 78.3 degrees F. Could someone show me how to do this?

SAMPLE STATISTICS: mean home temperature setting (x bar)=77.5°F, size (n)=40 homes.

POPULATION STATISTIC: standard deviation (σ)=1.9°F

CLAIM: Mean home temperature setting is 78.3°F.

COUNTERCLAIM: Mean home temperature setting is not 78.3°F.

SIGNIFICANCE LEVEL: ɑ=0.05 (95% confidence level)

Null hypothesis, H₀: µ=78.3°F(claim)

Alternative hypothesis, H₁: µ≠78.3°F (counterclaim)

This is a 2-tail test, because “not equals” means less than (left tail) or greater than (right tail), so the significance level is split between the two tails giving us the required confidence interval of 95%.

TEST STATISTIC: Z=(x bar-µ)/(σ/√n)=(77.5-78.3)/(1.9/√40)=-2.66, corresponding to a P-value of 0.0039.

ɑ/2=0.025 and sample P-value 0.0039<0.025, which means that H₀ is rejected, and H₁ is accepted to be true. The conclusion is the claim that the mean home temperature is not 78.3°F.

Another way of confirming this result is to use the 2-tail critical value for the 0.05 significance level, which corresponds to |Z|=1.96. -2.66 < -1.96, meaning that the sample Z-score is more extreme than the critical value, and H₀ is rejected and H₁ is accepted.

by Top Rated User (646k points)