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Chapter 10: Problem 4

The proportion of adults living in a small town who are college graduates is estimated to be \(\mathrm{p}=0.6\). To test this hypothesis, a random sample of 15 adults is selected. If the number of college graduates in our sample is anywhere from 6 to \(12,\) we shall not reject the null hypothesis that \(p=0.6 ;\) otherwise, we shall conclude that \(p \neq 0.6\) (a) Evaluate \(\alpha\) assuming that \(p=0.6\). Use the binomial distribution. (b) Evaluate \(\beta\) for the alternatives \(p=0.5\) and \(p-0.7\). (c) Is this a good test procedure?

Short Answer

Expert verified
Firstly, calculate the significance level (\(\alpha\)) using cumulative binomial distribution under the assumption \(p=0.6\). Then calculate the probabilities of making type II error (\(\beta\)) for \(p=0.5\) and \(p=0.7\). Lastly, evaluate the goodness of the test procedure based on \(\alpha\) and \(\beta\).

Step by step solution

01

Calculate Significance Level (\(\alpha\))

Firstly, we have to calculate the probability that the number of college graduates in our sample is either less than 6 or more than 12, under the assumption that \(p=0.6\). This can be done using the cumulative binomial distribution:\(\alpha = P(X<6) + P(X>12)\) where \(X\) is a random variable that follows the binomial distribution.
02

Calculate Type II Error (\(\beta\))

Next, we have to calculate the type II error, which is the probability of not rejecting the null hypothesis when it is false. This is done for the alternative values \(p=0.5\) and \(p=0.7\). This can be calculated using cumulative binomial distribution similar to \(\alpha\). The probability of making a type II error when \(p=0.5\) is:\(\beta = P(6 \leq X \leq 12)\) The probability of making a type II error when \(p=0.7\) is:\(\beta = P(6 \leq X \leq 12)\) where \(X\) is a random variable that follows the binomial distribution.
03

Evaluate the test procedure

Finally, the goodness of the test procedure is evaluated. A good test procedure is one that has low probabilities of making type I and type II errors. So, if the calculated \(\alpha\) and \(\beta\) are low, this would imply that the test procedure is good.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Type I Error
In hypothesis testing, when you mistakenly reject a true null hypothesis, you commit a Type I error. It's often called a "false positive." For example, you conclude that more or less than 60% of adults are college graduates when the actual proportion truly is 60%.
  • Imagine you're doing a medical test: a Type I error is like diagnosing a healthy person with a disease.
  • The probability of making this error is called the significance level, denoted as \(\alpha\).
For our small-town college graduate situation, to find \(\alpha\), you calculate how likely it is to get an extreme sample result assuming the null hypothesis is true. Here, the extremes are fewer than 6 or more than 12 college graduates out of 15. Lowering the significance level reduces the chance of a Type I error, but can increase the likelihood of other errors.
Type II Error
While a Type I error is about rejecting a true hypothesis, a Type II error is all about missing a false hypothesis. It's when you fail to reject a null hypothesis that is actually false. This is also known as a "false negative."
  • Continuing with the medical analogy, it's like giving a clean bill of health to someone who is actually ill.
  • The probability of this occurring is represented by \(\beta\).
In the exercise about the small town, you calculate \(\beta\) to know how likely it is to not detect the real proportion when it is 0.5 or 0.7, yet decide to accept the false 0.6 proposition. To lower \(\beta\), you can increase the sample size, which provides more data and often leads to more reliable testing outcomes.
Binomial Distribution
The binomial distribution is a probability distribution that summarizes the likelihood of a value taking one of two independent states. It鈥檚 a key player in hypothesis testing for scenarios like the one we're dealing with here.
  • The two possible outcomes in our example are: a person is a college graduate or not.
  • The binomial probability formula is \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), where \(n\) is the number of trials, \(k\) is the number of successes, and \(p\) is the probability of success on a single trial.
In the context of the exercise, we use this distribution to calculate both the probability of Type I and Type II errors. It helps in identifying the distribution of outcomes when sampling with a known probability, \(p\), providing a model that assumes each individual outcome is independent of others.
Significance Level
The significance level, \(\alpha\), is the threshold for determining whether a hypothesis test result is significant. It's like a cut-off point that tells us when to reject the null hypothesis.
  • Common significance levels include 0.05, 0.01, or 0.10.
  • If the probability of the observed outcome is below \(\alpha\), the result is considered statistically significant.
For our problem, setting \(\alpha\) would involve deciding the risk you're willing to take for a Type I error. An \(\alpha\) of 0.05 suggests you're prepared to wrongly reject a true hypothesis 5% of the time. This trade-off needs consideration alongside the potential for Type II errors, reflecting a balance between avoiding false positives and false negatives.

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Most popular questions from this chapter

10.35 To find out whether a new serum will arrest leukemia, 9 mice, all with an advanced stage of the disease, are selected. Five mice receive the treatment and 4 do not. Survival times, in years, from the time the experiment commenced are as follows. $$\begin{array}{l|ccccc}\text { Treatment } & 2.1 & 5.3 & 1.4 & 4.6 & 0.9 \\\\\hline \text { No Treatment } & 1.9 & 0.5 & 2.8 & 3.1 &\end{array}$$ At the 0.05 level of significance can the serum be said to be effective? Assume the two distributions to be normally distributed with equal variances.

In a study on the fertility of married women conducted by Martin O'Connell and Carolyn C. Rogers for the Census Bureau in 1979 , two groups of childless wives aged 25 to 29 were selected at random and each wife was asked if she eventually planned to have a child. One group was selected from among those wives married less than two years and the other from among those wives married five years. Suppose that 240 of 300 wives married less than two years planned to have children some day compared to 288 of the 400 wives married five years, Can we conclude that the proportion of wives married less than two years who planned to have children is significantly higher than the proportion of wives married five years? Make use of a P-value

10.36 \text { A large automobile manufacturing company is } trying to decide whether to purchase brand \(A\) or brand \(B\) tires for its new models. To help arrive at a decision, an experiment is conducted using 12 of each brand. The tires are run until they wear out. The results are $$\begin{aligned}\text { Brand } A: & x_{1}=37,900 \text { kilometers, } \\ & s_{1}=5,100 \text { kilometers. } \\\\\text { Brand B: } \quad x_{1} &=39,800 \text { kilometers, } \\\& s_{2}=5,900 \text { kilometers. }\end{aligned}$$ Test the hypothesis that there is no difference in the average wear of 2 brands of tires. Assume the populations to be approximately normally distributed with equal variances. Use a P-value.

An urban community would like to show that the incidence of breast cancer is higher than in a nearbyrural area. (PCB levels were found to be higher in the soil of the urban community.) If it is found that 20 of 200 adult women in the urban community have breast cancer and 10 of 150 adult women in the rural commus nity have breast cancer, can we conclude at the 0.05 level of significance that breast cancer is more prevalent in the urban community?

A fabric manufacturer believes that the proportion of orders for raw material arriving late is \(p=0.6\) If a random sample of 10 orders shows that 3 or fewer arrived late, the hypothesis that \(p=0.6\) should be rejected in favor of the alternative \(p<0.6\). Use the binomial distribution. (a) Find the probability of committing a type 1 error if the true proportion is \(p=0.6\) (b) Find the probability of committing a type II error for the alternatives \(p=0.3, p-0.4,\) and \(p=0.5\)
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