Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

M\&M'S Does Mars, Incorporated use the same proportion of red candies in its plain and peanut varieties? A random sample of 56 plain M\&M'S contained 12 red candies, and another random sample of 32 peanut M\&M'S contained 8 red candies. a. Construct a \(95 \%\) confidence interval for the difference in the proportions of red candies for the plain and peanut varieties. b. Based on the confidence interval in part a, can you conclude that there is a difference in the proportions of red candies for the plain and peanut varieties? Explain.

Short Answer

Expert verified
Answer: No, we cannot conclude that there is a significant difference in the proportions of red candies for the plain and peanut M&M'S varieties, since the confidence interval includes both positive and negative values, as well as zero.

Step by step solution

01

Calculate sample proportions and sample sizes

For the plain M&M'S: \(n_1 = 56\) and \(x_1 = 12,\) so the sample proportion is \(p_1 = \frac{x_1}{n_1} = \frac{12}{56}.\) For the peanut M&M'S: \(n_2 = 32\) and \(x_2 = 8,\) so the sample proportion is \(p_2 = \frac{x_2}{n_2} = \frac{8}{32}.\)
02

Calculate pooled proportion

\(p = \frac{(x_1 + x_2)}{(n_1 + n_2)} = \frac{(12 + 8)}{(56 + 32)}.\)
03

Calculate standard error

\(SE = \sqrt{p(1-p)(\frac{1}{n_1}+\frac{1}{n_2})}.\)
04

Find the critical value for a 95% confidence interval

The critical value for a \(95\%\) confidence interval is \(z^* = 1.96.\)
05

Calculate the margin of error

\(ME = z^* \times SE.\)
06

Calculate the confidence interval

The confidence interval will be \((p_1 - p_2) \pm ME.\) Now, we will calculate the confidence interval using the formulas and data above.
07

Calculate the confidence interval

\(p_1 = \frac{12}{56} = 0.2143.\) \(p_2 = \frac{8}{32} = 0.25.\) \(p = \frac{(12 + 8)}{(56 + 32)} = \frac{20}{88} = 0.2273.\) \(SE = \sqrt{0.2273(1-0.2273)(\frac{1}{56}+\frac{1}{32})} = 0.0902.\) \(ME = 1.96 \times 0.0902 = 0.17667.\) The confidence interval is \((0.2143 - 0.25) \pm 0.17667\) or \((-0.0357 \pm 0.17667),\) which, when calculated, results in the interval \((-0.21237, 0.14097).\)
08

Analyze the confidence interval

The confidence interval is \((-0.21237, 0.14097),\) which includes both positive and negative values. Since the interval contains zero, we cannot conclude that there is a significant difference in the proportions of red candies for the plain and peanut varieties. a. The 95% confidence interval for the difference in the proportions of red candies for the plain and peanut varieties is \((-0.21237, 0.14097).\) b. Based on the confidence interval in part a, we cannot conclude that there is a difference in the proportions of red candies for the plain and peanut varieties, since the interval includes both positive and negative values, as well as zero.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Sample Proportions
Sample proportions represent the fraction of items in a sample that possess a particular attribute. For example, if you're examining the proportion of red candies in a batch, the sample proportion would be the number of red candies divided by the total number of candies in the sample. Calculating the sample proportion is a fundamental step in many statistical analyses because it serves as an estimate of the population proportion. In hypothesis testing, sample proportions from different groups can be compared to infer whether there is a notable difference in the population.

When working with sample proportions, it's essential to consider sample sizes, as smaller samples are less likely to reflect the population accurately due to higher variability. Remember, larger sample sizes typically provide a more reliable estimate of the population proportion and reduce the margin of error in confidence interval calculations.
Pooled Proportion
In statistics, the pooled proportion is an estimate of the common proportion shared by two or more groups. This measure is used when we assume that the groups we're comparing all have the same underlying proportion for a characteristic of interest. To calculate it, we combine the number of successes from all groups and divide by the combined sample size.

A typical scenario that calls for a pooled proportion is when we're comparing two different treatments or conditions and want to know if they lead to the same result. The pooled proportion serves as the basis for calculating the standard error of the difference between two sample proportions, which is then used for constructing confidence intervals and carrying out hypothesis tests. It's crucial for the estimation of pooled proportions that the assumption of a common population proportion is reasonable; otherwise, the results may be misleading.
Standard Error
The standard error (SE) is a statistical measure that indicates the precision of an estimate. It is especially important when dealing with sample proportions, as it helps to quantify the variability of the estimate from the true population proportion. Think of the SE as an indication of how much you can expect the sample proportion to fluctuate from one sample to another if you repeated your sampling process.

In the context of confidence intervals, the SE is used to calculate the range within which we can be confident that the true population parameter lies. The formula for calculating the SE of a difference in sample proportions incorporates the pooled proportion when we assume the null hypothesis of no difference is true. A smaller SE suggests a more precise estimate, which in turn means a narrower confidence interval, providing more confidence in where the true population parameter may lie.
Margin of Error
The margin of error (ME) is a statistic that reflects the range of uncertainty around a sample estimate. It is used to create a confidence interval, which is a range of values that is likely to contain the population parameter with a certain level of confidence. The ME is calculated by multiplying the standard error by a critical value from the distribution (often a z-score corresponding to the desired level of confidence).

A larger ME indicates more uncertainty in the estimate, and thus a wider confidence interval. The size of the ME is influenced by factors such as the size of the sample, the variability within the data, and the level of confidence desired. In practice, a smaller margin of error is preferable as it suggests a more precise estimate of the population parameter. In the context of hypothesis testing, we use the ME to determine if the confidence interval includes zero, which helps us make decisions about the hypotheses.
Hypothesis Testing
Hypothesis testing is a formal methodology used in statistics to determine whether there is sufficient evidence to reject a null hypothesis. The process involves several steps, starting with the formation of two competing hypotheses: the null hypothesis, which usually suggests no effect or no difference, and the alternative hypothesis, which suggests some effect or difference.

In the case of comparing sample proportions, we calculate a test statistic and compare it to a critical value, or we construct a confidence interval. If a confidence interval does not contain the value of zero (for differences), we can say there is a statistically significant difference between the groups under consideration. On the other hand, if the confidence interval does include zero, the evidence is not strong enough to reject the null hypothesis, indicating no significant difference. This is seen in analysis of confidence intervals for proportions of red candies in M&M'S varieties—the interval included zero, so we didn't find significant evidence to suggest the proportions differ.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In an attempt to reduce their carbon footprint, many consumers are purchasing hybrid, plug-in hybrid, or electric cars. Consumer Reports ranks the Chevrolet Bolt first among electric cars, with an EPA rating of 238 miles between battery charges, although others report a range between 190 and 313 miles! \(!^{10}\) To test this claim, suppose that \(n=35\) road tests were conducted and that the average miles between charges was 232 miles with a standard deviation of 20.2 miles. a. Construct a \(95 \%\) confidence interval for \(\mu,\) the average time between battery charges for the Chevrolet Bolt. b. Does the confidence interval in part a confirm the EPA's claim of 238 miles per battery charge? Why or why not?

Calculate the margin of error in estimating a population mean \(\mu\) for the values given in Exercises \(7-10 .\) Comment on how an increased sample size affects the margin of error: \(n=100, s^{2}=4\)

Construct a \(95 \%\) confidence interval for the difference in the population means. Then find a point estimate for the difference in the population means and calculate the margin of error: Compare your results. Can you conclude that there is a difference in the two population means? $$\begin{aligned}&\text { } n_{1}=n_{2}=50, \bar{x}_{1}=125.2, \bar{x}_{2}=123.7, s_{1}=5.6\\\ &s_{2}=6.8\end{aligned}$$

Using the sample information given in Exercises \(22-23,\) give the best point estimate for the binomial proportion \(p\) and calculate the margin of error. A random sample of \(n=900\) observations from a binomial population produced \(x=655\) successes.

Calculate the margin of error in estimating a binomial proportion \(p\) for the sample sizes given in Exercises \(11-14\). Use \(p=.5\) to calculate the standard error of the estimator, and comment on how an increased sample size affects the margin of error. \(n=100\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free