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Find the proportions \(\hat{p}\) and \(\hat{q}\) for each. a. \(n=52, X=32\) b. \(n=80, X=66\) c. \(n=36, X=12\) d. \(n=42, X=7\) e. \(n=160, X=50\)

Short Answer

Expert verified
a. \(\hat{p} = 0.615\), \(\hat{q} = 0.385\); b. \(\hat{p} = 0.825\), \(\hat{q} = 0.175\); c. \(\hat{p} = 0.333\), \(\hat{q} = 0.667\); d. \(\hat{p} = 0.167\), \(\hat{q} = 0.833\); e. \(\hat{p} = 0.3125\), \(\hat{q} = 0.6875\).

Step by step solution

01

Define the Proportion 3

The sample proportion 03 is defined as the ratio of the number of successes \(X\) to the total number of trials \(n\). It is given by the formula \(\hat{p} = \frac{X}{n}\).
02

Compute 03 and 0f for a

For the first case where \(n=52\) and \(X=32\), substitute these values into the formula for \(\hat{p}\): \[\hat{p} = \frac{32}{52} = 0.615\]. Then, calculate \(\hat{q}\) using \(\hat{q} = 1 - \hat{p}\): \[\hat{q} = 1 - 0.615 = 0.385\].
03

Compute 03 and 0f for b

For \(n=80\) and \(X=66\), substitute into the formula for \(\hat{p}\): \[\hat{p} = \frac{66}{80} = 0.825\]. Then, \(\hat{q}\) is \(\hat{q} = 1 - \hat{p}\): \[\hat{q} = 1 - 0.825 = 0.175\].
04

Compute 03 and 0f for c

For \(n=36\) and \(X=12\), compute \(\hat{p}\) as: \[\hat{p} = \frac{12}{36} = 0.333\]. Then find \(\hat{q}\): \[\hat{q} = 1 - 0.333 = 0.667\].
05

Compute 03 and 0f for d

For \(n=42\) and \(X=7\), calculate \(\hat{p}\): \[\hat{p} = \frac{7}{42} = 0.167\]. Then, \(\hat{q}\): \[\hat{q} = 1 - 0.167 = 0.833\].
06

Compute 03 and 0f for e

For \(n=160\) and \(X=50\), determine \(\hat{p}\): \[\hat{p} = \frac{50}{160} = 0.3125\]. Then compute \(\hat{q}\): \[\hat{q} = 1 - 0.3125 = 0.6875\].

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Key Concepts

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

Proportion Calculation
In statistics, a proportion is a special type of ratio that compares a part to a whole. It is often expressed as a fraction or a percentage. This is frequently used in scenarios where you want to understand how a part contributes to the entire set. For instance, if you're measuring how many students in a class of 100 like math, and 75 of them do, the proportion would be 75/100.

When we calculate proportions, the key components are the number of successes (or what you're counting) and the total number of trials (or the whole group size). The formula \(ackslash{hat{p} = \frac{X}{n}}\) where \(X\) is the count of your interest (success) and \(n\) is the total size of your group (trials), allows us to easily find the relative size of the part compared to the whole.
Sample Proportions
A sample proportion, represented as \(\hat{p}\), is a way to estimate the proportion from a larger population based on a smaller, representative sample. Imagine you want to know how popular a new movie is, but you only ask 50 people. If 30 like it, \(\hat{p}\) would be \(\frac{30}{50} = 0.6\) or 60%. This sample proportion helps in predicting the likely proportion for the entire population.

Sample proportions are central to inferential statistics because they allow us to make guesses about the larger population without needing information from everyone in it. This is crucial because it saves time and resources, while still providing a data-supported estimate. When you calculate a sample proportion, also remember \(\hat{q} = 1 - \hat{p}\) for the complement, which can be useful in understanding the rest of the dataset.
Statistical Notation
Statistical notation plays an essential role in the communication of quantitative data analyses. Notations like \(\hat{p}\) and \(n\) help keep our work precise and easy to follow. In context, \(\hat{p}\) denotes the sample proportion of successes, while \(n\) is the total number of observations or trials. Another important symbol is \(X\), which is the number of successes in your sample.

Using standardized notation is crucial because it makes mathematical expressions universally recognizable and readable. It avoids confusion when sharing findings with others, allowing for a seamless exchange of information in the statistical community. When diving into more complex topics, solid understanding of these notations will build your confidence and accuracy in problem-solving.
Successes and Trials
Successes and trials are terms used in binomial experiments or processes where the outcome is either a success or a failure. Here, a 'success' doesn't mean a positive outcome but the event or result we're interested in counting. The number of successes \(X\) and the number of trials \(n\) are the primary components to determine proportions.

In situations where you're calculating proportions, understanding what constitutes a success is crucial. If you surveyed 80 people, and 55 prefer chocolate ice cream (X), then a success would be each person saying they like chocolate, making \(X=55\). When these counts are plugged into the sample proportion formula, they allow us to derive insights into behavior or tendencies from specific populations. Understanding successes and trials will also aid in grasping advanced probability concepts later.

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Most popular questions from this chapter

Commuting Times The U.S. Census Bureau reports that the average commuting time for citizens of both Baltimore, Maryland, and Miami, Florida, is approximately 29 minutes. To see if their commuting times appear to be any different in the winter, random samples of 40 drivers were surveyed in each city and the average commuting time for the month of January was calculated for both cities. The results are shown. At the 0.05 level of significance, can it be concluded that the commuting times are different in the winter? $$ \begin{array}{lcc}{} & {\text { Miami }} & {\text { Baltimore }} \\ \hline \text { Sample size } & {40} & {40} \\ {\text { Sample mean }} & {28.5 \min } & {35.2 \mathrm{min}} \\ {\text { Population standard deviation }} & {7.2 \mathrm{min}} & {9.1 \mathrm{min}}\end{array} $$

For these exercises, perform each of these steps. Assume that all variables are normally or approximately normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified and assume the variances are unequal. Tax-Exempt Properties A tax collector wishes to see if the mean values of the tax-exempt properties are different for two cities. The values of the tax- exempt properties for the two random samples are shown. The data are given in millions of dollars. At \(\alpha=0.05,\) is there enough evidence to support the tax collector's claim that the means are different? $$ \begin{array}{cccc|cccc}{} & {\text { City } \mathbf{A}} & {} & {} & {} & {\text { City } \mathbf{B}} & {} & {} \\ \hline 113 & {22} & {14} & {8} & {82} & {11} & {5} & {15} \\ {25} & {23} & {23} & {30} & {295} & {50} & {12} & {9} \\ {44} & {11} & {19} & {7} & {12} & {68} & {81} & {2} \\ {31} & {19} & {5} & {2} & {} & {20} & {16} & {4} & {5}\end{array} $$

For Exercises 2 through \(12,\) perform each of these steps. Assume that all variables are normally or approximately normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Overweight Dogs A veterinary nutritionist developed a diet for overweight dogs. The total volume of food consumed remains the same, but one-half of the dog food is replaced with a low-calorie "filler" such as canned green beans. Six overweight dogs were randomly selected from her practice and were put on this program. Their initial weights were recorded, and they were weighed again after 4 weeks. At the 0.05 level of significance, can it be concluded that the dogs lost weight? $$ \begin{array}{l|cccccc}{\text { Before }} & {42} & {53} & {48} & {65} & {40} & {52} \\ \hline \text { After } & {39} & {45} & {40} & {58} & {42} & {47}\end{array} $$

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