Testing a Proportion: What Would Happen Under the Null? — Solutions and Rubrics

Let p be the proportion of all city residents who carry a reusable bottle. Use a one-sample z-test for a population proportion. H₀:p=0.40 is the assumed status quo; Hₐ:p>0.40 represents the investigator’s claim. Calculate the test distribution assuming the null value 0.40 is true; the sample proportion is not the parameter.

The SRS supplies randomization. For sampling without replacement, 200≤0.10(12,000)=1,200. The null expected successes are 200(0.40)=80 and failures 200(0.60)=120, both at least 10. Use p₀, because this is the sampling distribution being modeled under H₀, not an interval estimating an unknown p.

p̂=98/200=0.49. Under H₀ the standard deviation is √[0.40(0.60)/200]=0.0346410; z=(0.49−0.40)/0.0346410=2.59808. The null z distribution is approximately standard normal because the count condition holds. For Hₐ:p>0.40, p-value=P(Z≥2.59808)≈0.004687, from the supplied table or a normal upper-tail command. Modeling the repeated-sample null distribution by this theoretical curve approximates the chance of an observed statistic at least this high.

If the population bottle-carrying proportion were 0.40, the probability of obtaining a sample proportion at least 0.49 in a random sample of 200 is approximately 0.004687. Since 0.004687≤0.05, reject H₀. There is convincing statistical evidence that more than 40% of city residents carry a reusable bottle. This is conditional on a fixed null parameter; it does not assign a probability of truth to H₀ and does not prove the alternative.

Upper count=8+66=74; simulated p=74/1,000=0.074. Assuming the machine’s red-token proportion is 0.50, about 7.4% of simulated samples of 400 had p̂ at least 0.5375. The 859 central results are not as extreme in the greater direction; equality is included.

Lower count=61+6=67; simulated p=0.067. The theoretical probability P(Z≤−1.5)=0.0668072 is close but need not equal a finite simulation estimate. Both calculations condition on p=0.50 and use the lower alternative’s direction.

At least as extreme means z≤−|1.5| or z≥|1.5|. Include 61+6+8+66=141 simulations: p≈0.141. Theoretical P(Z≤−|1.5|)+P(Z≥|1.5|)=0.1336144. The simulated lower and upper counts are 67 and 74, so doubling 0.074 gives 0.148 rather than the actual 0.141; finite simulated tails need not be exactly symmetric.

First maker: p is its population proportion of packages arriving within two days. H₀:p≥0.60 expresses its claim and Hₐ:p<0.60 expresses the consumer concern. Test at the boundary H₀:p=0.60. For other questions, Hₐ:p>0.60 is one-sided and Hₐ:p≠0.60 is two-sided. Second maker: let p₂ be the proportion of all its packages arriving within two days. Its at-most claim is H₀:p₂≤0.70; the investigator’s alternative is Hₐ:p₂>0.70. The one-sided reference calculation uses the equality boundary H₀:p₂=0.70. Thus null claims can include =, ≥ or ≤, with equality included and the boundary tested. Every hypothesis concerns a population proportion, not the observed first-maker sample proportion 81/150.

Assuming p=0.60, about 6.681% of random samples of 150 would have an on-time sample proportion no greater than the observed 0.54. Because 0.06681>0.05, fail to reject H₀; there is not convincing statistical evidence that fewer than 60% arrive within two days. A non-small p-value indicates the result is not unusual enough under H₀ at this threshold. It is neither the probability the maker is right nor evidence proving H₀. A smaller p-value makes the observed result less compatible with the null model and supplies stronger evidence for the lower alternative.

A passes randomization and 120≤800 but fails normality: expected successes 120(0.02)=2.4<10, despite 117.6 expected failures. B has adequate expected counts 250 and 250, but volunteers do not establish randomization; unknown population/sampling design cannot establish the without-replacement condition. C has randomization and 90 expected successes/failures, but 180>100 violates the 10% condition. More observations do not make a voluntary sample random or remove the substantial sampling fraction, and rare-success counts must be checked directly.

Alpha is the predetermined chance of rejecting H₀ when H₀ is true, over repeated valid tests. It may be specified by the problem or chosen by the researcher before looking at results. At 0.01, 0.032>0.01: fail to reject. At 0.05, 0.032≤0.05: reject. The observed p-value is the same; the decision threshold differs. Choosing a favorable threshold after seeing results does not follow the procedure whose preset Type I rate was stated.