/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 A certain type of flashlight is ... [FREE SOLUTION] | 91Ó°ÊÓ

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A certain type of flashlight is sold with the four batteries included. A random sample of 150 flashlights is obtained, and the number of defective batteries in each is determined, resulting in the following data: \begin{tabular}{l|ccccc} Number Defective & 0 & 1 & 2 & 3 & 4 \\ \hline Frequency & 26 & 51 & 47 & 16 & 10 \end{tabular} Let \(X\) be the number of defective batteries in a randomly selected flashlight. Test the null hypothesis that the distribution of \(X\) is \(\operatorname{Bin}(4, \theta)\). That is, with \(p_{i}=P(i\) defectives), test \(H_{0}: p_{i}=\left(\begin{array}{l}4 \\ i\end{array}\right) \theta^{i}(1-\theta)^{4-i} \quad i=0,1,2,3,4\) [Hint: To obtain the mle of \(\theta\), write the likelihood (the function to be maximized) as \(\theta^{u}(1-\theta)^{v}\), where the exponents \(u\) and \(v\) are linear functions of the cell counts. Then take the natural \(\log\), differentiate with respect to \(\theta\), equate the result to 0 , and solve for \(\hat{\theta}\).]

Short Answer

Expert verified
MLE for \(\theta\) is 0.3083; perform a chi-square test to validate the null hypothesis.

Step by step solution

01

Understand the Problem

We need to test if the number of defective batteries follows a binomial distribution \( \operatorname{Bin}(4, \theta) \), where each flashlight has 4 batteries. We will test the null hypothesis that the distribution follows this binomial form.
02

Determine Total Defective Batteries

First, calculate the total number of defective batteries in the sample. Sum the products of the number of defectives and their frequencies: \( 0\times26 + 1\times51 + 2\times47 + 3\times16 + 4\times10 = 185 \).
03

Calculate MLE for \(\theta\)

We know \( X \) follows \( \operatorname{Bin}(4, \theta) \). The likelihood function is expressed as \( L(\theta) = \theta^{u}(1-\theta)^{v} \). Here, \( u = 185 \), the total number of defective batteries, and \( v = 4\times150 - u = 600 - 185 = 415 \). Maximize this to find the MLE: \( \hat{\theta} = \frac{u}{4 \times 150} = \frac{185}{600} = 0.3083 \).
04

Establish the Null Hypothesis

The null hypothesis is that the probability of \( i \) defectives follows \( p_{i}=\binom{4}{i} \theta^{i}(1-\theta)^{4-i} \). We will use \( \hat{\theta} = 0.3083 \) to test this.
05

Calculate Binomial Probabilities

Using the binomial formula: \( p_{i}=\binom{4}{i} (0.3083)^{i} (1-0.3083)^{4-i} \) compute the probabilities \( P(0) - P(4) \) for 0 to 4 defectives.
06

Expected Frequencies Calculation

Multiply each \( p_{i} \) by 150 (the sample size) to get expected frequencies for each number of defectives. This yields expected values for comparison with the observed frequencies.
07

Perform Chi-Square Test

Calculate the chi-square statistic: \( \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \), where \( O_i \) are observed and \( E_i \) are expected frequencies. Compare \( \chi^2 \) to the critical value from \( \chi^2 \) distribution with 4-1=3 degrees of freedom at an appropriate significance level.
08

Draw Conclusion

If the chi-square statistic exceeds the critical value, reject \( H_{0} \). Otherwise, there's not enough evidence to reject \( H_{0} \), suggesting that the data is consistent with a binomial distribution.

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Key Concepts

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

Maximum Likelihood Estimation
Maximum Likelihood Estimation (MLE) is a method used to estimate the parameters of a statistical model. When the model is binomial, as in this case, we use MLE to find the probability parameter, \( \theta \), which represents the chance of a single battery being defective in a flashlight. The likelihood function shows how likely the observed data would be, given a particular value of \( \theta \). For our problem, it can be simplified to \( \theta^u (1-\theta)^v \), where \( u \) is the total number of defective batteries, and \( v \) is the total number of non-defective batteries. The goal of MLE is to find the value of \( \theta \) that maximizes this likelihood function. By taking the natural logarithm of the likelihood (for easier computation), differentiating with respect to \( \theta \), and setting the resulting equation to zero, we solve for \( \hat{\theta} \). In our example, this leads to \( \hat{\theta} = \frac{185}{600} = 0.3083 \). This value of \( \theta \) is the maximum likelihood estimate and indicates that approximately 30.83% of the batteries are defective.
Chi-Square Test
The Chi-Square Test is a statistical method used to compare observed data with data we would expect to obtain based on a particular hypothesis. It helps to determine if there's a significant difference between expected and observed frequencies. In our exercise, we use the Chi-Square Test to check if the distribution of defective batteries follows a binomial distribution with the estimated \( \hat{\theta} = 0.3083 \). To perform the test, we first calculate the expected frequencies for each number of defective batteries using the binomial formula. Each probability \( p_i \) is then multiplied by the sample size (150) to find the expected number of batteries for each category.The Chi-Square statistic is computed as \( \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \), where \( O_i \) and \( E_i \) represent observed and expected frequencies, respectively. The resulting \( \chi^2 \) value is compared against a critical value from the chi-square distribution table with 3 degrees of freedom (since we have four possible outcomes and \( k-1 \) degrees of freedom). If the \( \chi^2 \) statistic is greater than the critical value, the null hypothesis that the data follows the binomial distribution is rejected.
Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions about population parameters based on sample data. In our exercise, the null hypothesis \( H_0 \) claims that the number of defective batteries follows a binomial distribution \( \text{Bin}(4, \theta) \). First, we establish our null hypothesis, which states that the probability \( p_i \) of having \( i \) defective batteries is given by the binomial probabilities \( \binom{4}{i} \theta^i (1-\theta)^{4-i} \) using the MLE \( \hat{\theta} = 0.3083 \). The alternative hypothesis \( H_a \) would suggest that the distribution does not fit this model.We use the Chi-Square Test to evaluate the hypothesis. The calculated \( \chi^2 \) statistic is compared with a critical value that is based on our chosen significance level (commonly 0.05). A \( \chi^2 \) statistic that is larger than the critical value would lead us to reject the null hypothesis, indicating our sample does not match a binomial distribution. Conversely, if the \( \chi^2 \) statistic is smaller, it supports the null hypothesis, suggesting the observed data does fit a binomial distribution.
Defective Items Analysis
Defective Items Analysis involves examining items to determine the proportion that are faulty. In our problem, we focus on analyzing defective batteries within flashlights. Such analyses are crucial for quality control in manufacturing processes. In this exercise, we gathered data from a sample of 150 flashlights, each containing four batteries. A breakdown of frequencies shows how many flashlights had 0, 1, 2, 3, or 4 defective batteries. This provides valuable insights into the quality of the flashlights being produced. Analyzing defective items helps companies to improve production processes and maintain quality standards. If the number of defects is higher than expected, it may indicate problems in the production line or sourcing of materials. This analysis lays the groundwork for addressing and solving such issues by providing a quantitative base for decision-making and potential adjustments in production methods.

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Most popular questions from this chapter

The NCAA basketball tournament begins with 64 teams that are apportioned into four regional tournaments, each involving 16 teams. The 16 teams in each region are then ranked (seeded) from 1 to 16. During the 12-year period from 1991 to 2002 , the top-ranked team won its regional tournament 22 times, the second-ranked team won 10 times, the third-ranked team won 5 times, and the remaining 11 regional tournaments were won by teams ranked lower than 3 . Let \(P_{i j}\) denote the probability that the team ranked \(i\) in its region is victorious in its game against the team ranked \(j\). Once the \(P_{i j}\) 's are available, it is possible to compute the probability that any particular seed wins its regional tournament (a complicated calculation because the number of outcomes in the sample space is quite large). The paper "Probability Models for the NCAA Regional Basketball Tournaments"(Amer. Statist., 1991: 35-38) proposed several different models for the \(P_{i j}\) 's. a. One model postulated \(P_{i j}=.5-\lambda(i-j)\) with \(\lambda=\frac{1}{32}\) (from which \(P_{16,1}=\frac{1}{32}, P_{16,2}=\frac{2}{32}\), etc.). Based on this, \(P\) (seed #1 wins) \(=.27477\), \(P(\) seed \(\\# 2\) wins \()=.20834\), and \(P\) (seed #3 wins \()=.15429\). Does this model appear to provide a good fit to the data? b. A more sophisticated model has \(P_{i j}=.5+\) \(.2813625\left(z_{i}-z_{j}\right)\), where the \(z\) 's are measures of relative strengths related to standard normal percentiles [percentiles for successive highly seeded teams are closer together than is the case for teams seeded lower, and .2813625 ensures that the range of probabilities is the same as for the model in part (a)]. The resulting probabilities of seeds 1,2 , or 3 winning their regional tournaments are \(.45883, .18813\), and \(.11032\), respectively. Assess the fit of this model.

The article "A Probabilistic Analysis of Dissolved Oxygen-Biochemical Oxygen Demand Relationship in Streams" \((J\). Water 91Ó°ÊÓ Control Fed., 1969: 73-90) reports data on the rate of oxygenation in streams at \(20^{\circ} \mathrm{C}\) in a certain region. The sample mean and standard deviation were computed as \(\bar{x}=.173\) and \(s=.066\), respectively. Based on the accompanying frequency distribution, can it be concluded that oxygenation rate is a normally distributed variable? Use the chisquared test with \(\alpha=.05\). $$ \begin{array}{lc} \hline \text { Rate (per day) } & \text { Frequency } \\ \hline \text { Below .100 } & 12 \\ .100 \text {-below .150 } & 20 \\ .150 \text {-below .200 } & 23 \\ .200 \text {-below .250 } & 15 \\ .250 \text { or more } & 13 \\ \hline \end{array} $$

The article "Psychiatric and Alcoholic Admissions Do Not Occur Disproportionately Close to Patients' Birthdays"' (Psych. Rep., 1992: 944–946) focuses on the existence of any relationship between date of patient admission for treatment of alcoholism and patient's birthday. Assuming a 365day year (i.e., excluding leap year), in the absence of any relation, a patient's admission date is equally likely to be any one of the 365 possible days. The investigators established four different admission categories: (1) within 7 days of birthday, (2) between 8 and 30 days, inclusive, from the birthday, (3) between 31 and 90 days, inclusive, from the birthday, and (4) more than 90 days from the birthday. A sample of 200 patients gave observed frequencies of \(11,24,69\), and 96 for categories 1,2 , 3 , and 4, respectively. State and test the relevant hypotheses using a significance level of \(.01\).

The article "Compatibility of Outer and Fusible Interlining Fabrics in Tailored Garments (Textile Res. J., 1997: 137-142) gave the following observations on bending rigidity \((\mu \mathrm{N} \cdot \mathrm{m})\) for medium-quality fabric specimens, from which the accompanying MINITAB output was obtained: \(\begin{array}{rrrrrrrr}24.6 & 12.7 & 14.4 & 30.6 & 16.1 & 9.5 & 31.5 & 17.2 \\\ 46.9 & 68.3 & 30.8 & 116.7 & 39.5 & 73.8 & 80.6 & 20.3 \\ 25.8 & 30.9 & 39.2 & 36.8 & 46.6 & 15.6 & 32.3 & \end{array}\) Would you use a one-sample \(t\) confidence interval to estimate true average bending rigidity? Explain your reasoning.

A statistics department at a large university maintains a tutoring center for students in its introductory service courses. The center has been staffed with the expectation that \(40 \%\) of its clients would be from the business statistics course, \(30 \%\) from engineering statistics, \(20 \%\) from the statistics course for social science students, and the other \(10 \%\) from the course for agriculture students. A random sample of \(n=120\) clients revealed \(52,38,21\), and 9 from the four courses. Does this data suggest that the percentages on which staffing was based are not correct? State and test the relevant hypotheses using \(\alpha=.05\).

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