One Mean and Paired Differences: A Complete t Investigation — Solutions and Rubrics

μ is the mean longevity in hours for the focus-group population using this perfume. With unknown population σ, use a one-sample t-test and one-sample t-interval, estimating σ by sample s. H₀:μ=12; Hₐ:μ≠12 for “differs.” The one-sided alternatives would be μ<12 or μ>12; μ₀=12 is the null value. Student t accounts for estimating variability rather than treating s as known population σ.

The SRS establishes randomization. Sampling without replacement requires 16≤0.10(4,000)=400. Sorted values range from 11.14 to 14.01 and cluster around 12.5; no isolated extreme or strong skew is evident. Using median-of-halves quartiles gives Q₁=(11.83+12.05)/2=11.94 and Q₃=(13.25+13.33)/2=13.29, IQR=1.35, fences 9.915 and 15.315; no observations lie outside them. Since n=16<30 and no population Normality was given, inspecting the sample distribution supports the small-sample condition. The same observations/design justify the test and interval; n alone is insufficient.

x̄=12.569375 hours; s=0.8597090 hours; SE=s/√16=0.2149273 hours. t=(12.569375−12)/0.2149273=2.6491522; df=15. Under H₀ the test statistic has a t₁₅ distribution, so p=P(T≤−2.6491522)+P(T≥2.6491522)=2(0.00911110)=0.0182222, found using the supplied t tail or technology. If the population mean longevity were 12 hours, about 1.82% of such random samples would have a standardized sample mean at least this far from 12 in either direction.

With df=15 and central 95% t critical 2.1314495, MOE=2.1314495(0.2149273)=0.4581066 hours. The interval is 12.569375±0.4581066=(12.1112684,13.0274816) hours. We are 95% confident that the mean longevity for this focus-group population lies between about 12.111 and 13.027 hours. In repeated random sampling from the same population, about 95% of intervals constructed by this method capture the population mean. This one computed interval either contains μ or does not; 95% is not a probability that μ varies. The interval excludes 12, supporting a different mean and agreeing with the two-sided 5% test.

0.0182222≤0.05, so reject H₀:μ=12. There is convincing statistical evidence that the mean longevity of this perfume for the focus-group population differs from 12 hours. The conclusion follows the two-sided alternative, uses non-definitive evidence language, and does not claim every person experiences more than 12 hours. Both tails matter because differences below and above the target are relevant; the upper tail alone is half the p-value.

After−before differences are 1,2,2,3,3,4,4,5,5,6,6,7 points. Their mean is 4 points and sample s_d=1.8586408 points. μ_d is the population mean after-minus-before score change among enrolled learners. Use a one-sample t-interval/test on the single sample of differences, with n=12 pairs. Measurements within a learner are dependent; independent two-sample methods ignore the pairing and would estimate the wrong sampling variability.

H₀:μ_d=0 and Hₐ:μ_d≠0; an increase question uses μ_d>0 and a decrease uses μ_d<0. Learners were randomly selected. Twelve pairs satisfy 12≤0.10(2,400)=240. The difference sample is symmetric about 4, with no isolated extreme values: its frequencies are 1:1,2:2,3:2,4:2,5:2,6:2,7:1. With 12<30 pairs, this sample of differences, not the before/after marginal distributions, supports the shape condition. These checks justify both one-sample t procedures for μ_d.

SE=s_d/√12=0.5365434 points; t=(4−0)/0.5365434=7.4551289, df=11. p=2(0.00000634549)=0.00001269099. If the population mean after-minus-before change were zero, a random sample of 12 learners would have a standardized mean difference at least 7.4551 in absolute value with probability about 0.00001269. Since p≤0.05, reject H₀; there is convincing evidence of a population mean change for enrolled learners. Pairing alone does not eliminate time/practice confounding.

MOE=2.2009852(0.5365434)=1.1809240 points. The interval is 4±1.1809240=(2.8190760,5.1809240) points. We are 95% confident the enrolled-learner population mean after-minus-before change is between about 2.819 and 5.181 points. In repeated SRSs of 12 from this same population, about 95% of such intervals capture μ_d. This fixed interval either captures μ_d or does not. Its entirely positive values support an increase in population mean score, not an increase for every individual or a proven instructional effect.

All studies use an SRS and sample no more than 40≤300 from their populations. A has n=8 but a Normal population, one route to the condition. B has n=40≥30, satisfying the course large-sample route; extreme observations would still warrant practical inspection. C has 40 differences≥30; n counts pairs/differences, not 80 observations. D has only eight differences and an isolated 30, showing a strong upper extreme, so the small-sample difference distribution does not justify t inference. Without a stated Normal population or at least 30 values/differences, inspect the relevant sample for strong skewness and outliers. Random selection does not remove D’s shape problem.

Analyze the one sample of ten within-pair treatment differences with a one-sample t procedure. Random assignment within pairs supplies the experiment’s randomization. Inspect the ten differences; they are roughly balanced with no isolated extreme, supporting the small-sample condition. The 10% condition is for sampling without replacement, not randomized treatment assignment, so there is no population-sampling fraction to verify here. The experiment can support a causal treatment comparison under the design, while volunteer participation limits generalization to a larger population.

Student t distributions form a symmetric, bell-shaped, standardized family indexed by degrees of freedom based on sample size. With small df they have heavier tails and a lower peak than the standard normal; as df increases they approach the standard normal with mean 0 and SD 1. Unknown population σ is estimated by sample s, adding uncertainty represented by t rather than assuming s is a known population SD. For the one-sample procedures df=n−1, so here df=35.

SE=12/√36=2 hours. At 90%, MOE=1.68957246(2)=3.37914492, width=6.75828983. At 95%, MOE=2.03010793(2)=4.06021586, width=8.12043171. For this fixed sample, higher confidence increases critical value, MOE and width to capture more of the theoretical sampling distribution.

At n=144, SE=12/√144=1 hour, MOE=1.97669220, width=3.95338440. The width ratio is 0.48684412. Quadrupling n halves SE and approximately halves width at fixed confidence and variability. Exact width differs from one half because df increases and t* decreases; width is approximately proportional to 1/√n, not 1/n.

Assuming the population mean battery life is 36 hours, the probability of a standardized sample mean at least as low as observed in a sample of the same size is 0.22. Since 0.22>0.05, fail to reject H₀; there is not convincing evidence that the customer-use population mean is less than 36 hours. An outcome that is not unusual enough under the null does not prove the advertised mean or exclude smaller means; sample uncertainty and power remain.

We are 95% confident that the customer-use population mean battery life lies from 34.8 (lower) to 37.0 (upper) hours. In repeated sampling with the same method/size/population, about 95% of constructed intervals contain the fixed mean; this computed interval either does or does not. The value 36 is plausible within the interval, so it does not provide convincing two-sided evidence against 36. It is an interval for a mean, not for each battery’s lifetime. A central 95% interval matches a two-sided 5% test; the one-sided question uses a different tail allocation, so matching requires care.