/*! 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 57 Accept a credit card? A bank wan... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Accept a credit card? A bank wants to estimate the proportion of people who would agree to take a credit card they offer if they send a particular mailing advertising it. For a trial mailing to a random sample of 100 potential customers, 0 people accept the offer. Can they conclude that fewer than \(10 \%\) of their population of potential customers would take the credit card? Answer by finding an appropriate \(95 \%\) confidence interval.

Short Answer

Expert verified
Yes, fewer than 10% would accept the credit card offer.

Step by step solution

01

Identify the Problem

We need to determine whether fewer than \(10\%\) of potential customers would accept the credit card offer. This requires constructing a confidence interval for the proportion of people who respond positively to the offer.
02

Define the Sample Proportion

Given that in a sample of 100 potential customers, 0 accepted the offer, the sample proportion \(\hat{p}\) is \( \frac{0}{100} = 0 \).
03

Calculate the Standard Error

The standard error (SE) for the proportion is calculated using the formula \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \(\hat{p} = 0\) and \(n = 100\). So, \( SE = \sqrt{\frac{0 \times (1-0)}{100}} = 0 \).
04

Compute the Confidence Interval

The confidence interval is given by \( \hat{p} \pm Z \times SE \) where \( Z \) is the critical value from the standard normal distribution. For a \(95\%\) confidence level, \( Z \approx 1.96 \). Therefore, the interval is \( 0 \pm 1.96 \times 0 = 0 \).
05

Interpret the Results

The confidence interval we computed is \(0\) which does not cover \(10\%\). This suggests that fewer than \(10\%\) of the population would accept the credit card offer.

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 91Ó°ÊÓ!

Key Concepts

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

Sample Proportion
When estimating a proportion of a population who might respond positively to an offer or display a certain behavior, the sample proportion is crucial. It is denoted by \( \hat{p} \). This value represents the fraction of the sample showing the desired characteristic. In our scenario, the bank initially targets 100 individuals. But since none responded affirmatively to the credit card offer, their sample proportion is 0.Understanding the sample proportion helps illustrate what percentage of the surveyed group acts a certain way. If one or two people had accepted out of the 100, the sample proportion would have been \( \frac{1}{100} \) or \( \frac{2}{100} \), illustrating that even small numbers can change the perception of acceptability in different contexts.
Standard Error
The Standard Error (SE) is fundamental in constructing a confidence interval. It measures how much variability there is in a sample proportion and indicates how much our \(\hat{p}\) might differ from the real population proportion \(p\). It gets smaller as the sample size increases.To calculate SE for a proportion, use the following formula: \\[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]Here, \(\hat{p}\) is 0 for our situation because no one accepted the offer, and \(n\) is 100.With this formula:
  • \( \hat{p} = 0 \)
  • \( 1-\hat{p} = 1 \)
  • \( n = 100 \)
Therefore, our standard error is \( \sqrt{\frac{0 \times 1}{100}} = 0 \).Having the SE as zero meant the variability in reading acceptance was minimal in this sample.
Critical Value
The Critical Value aids in determining the range of values that are plausible estimates of the population parameter at our selected confidence level. It's linked to the confidence level, acting as a "cut-off" point for how extreme a result can be before it’s considered unlikely.In our case, with a \(95\%\) confidence level, we use the critical value from the standard normal distribution (Z-distribution), which commonly is \(1.96\). This represents the number of standard errors we move away from \(\hat{p}\) to develop a range that likely covers the true population parameter.Getting this critical value right ensures that our confidence interval provides a reasonable assurance regarding where the true proportion lies — crucial in determining whether our bank's assumptions about acceptance rates hold under statistical scrutiny.
Confidence Level
This is a statistical concept that offers us the faith or certainty with which we can state our interval estimate contains the true population parameter. In our problem, a \(95\%\) confidence level implies that if we could repeat our mail-out over numerous samples, approximately 95 out of 100 of those samples' intervals would contain the true proportion of advertisement acceptors.Choosing a \(95\%\) confidence level is standard as it strikes a balance between being too lenient and overly stringent. However, selecting a higher confidence level increases the interval's breadth, lowering our precision but improving security against missing the true parameter. On the flip side, a lower confidence level tightens our interval — meaning our estimation is more narrow, but also the risk increases that our true parameter lies outside this interval.

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

Multiple choice: CI property Increasing the confidence level causes the margin of error of a confidence interval to \((\) a) increase, \((b)\) decrease, \((c)\) stay the same.

Political views The General Social Survey asks respondents to rate their political views on a seven-point scale, where \(1=\) extremely liberal, \(4=\) moderate, and \(7=\) extremely conservative. A researcher analyzing data from the 2008 GSS obtains MINITAB output: a. Show how to construct the confidence interval from the other information provided. b. Can you conclude that the population mean is higher than the moderate score of \(4.0 ?\) Explain. c. Would the confidence interval be wider, or narrower, (i) if you constructed a \(99 \%\) confidence interval and (ii) if \(n=500\) instead of \(1933 ?\)

Fear of breast cancer A recent survey of 1000 American women between the ages of 45 and 64 asked them what medical condition they most feared. Of those sampled, \(61 \%\) said breast cancer, \(8 \%\) said heart disease, and the rest picked other conditions. By contrast, currently about \(3 \%\) of female deaths are due to breast cancer, whereas \(32 \%\) are due to heart disease. \(^{5}\) a. Construct a \(90 \%\) confidence interval for the population proportion of women who most feared breast cancer. Interpret. b. Indicate the assumptions you must make for the inference in part a to be valid.

Nutrient effect on growth rate Researchers are interested in the effect of a certain nutrient on the growth rate of plant seedlings. Using a hydroponics grow procedure that utilized water containing the nutrient, they planted six tomato plants and recorded the heights of each plant 14 days after germination. Those heights, measured in millimeters, were as follows: \(55.5,60.3,60.6,62.1,65.5,69.2 .\) a. Find a point estimate of the population mean height of this variety of seedling 14 days after germination. b. A method that we'll study in Section 8.3 provides a margin of error of \(4.9 \mathrm{~mm}\) for a \(95 \%\) confidence interval for the population mean height. Construct that interval. c. Use this example to explain why a point estimate alone is usually insufficient for statistical inference.

\(m\) and \(n\) Consider the sample size formula \(n=\left[\hat{p}(1-\hat{p}) z^{2}\right] / m^{2}\) for estimating a proportion. When \(\hat{p}\) is close to 0.50 , for \(95 \%\) confidence explain why this formula gives roughly \(n=1 / m^{2}\).

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.