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Website An investment company is planning to upgrade the mobile access to their website, but they'd like to know the proportion of their customers who access it from their smartphones. They draw a random sample of 200 from customers who recently logged in and check their IP address. Suppose that the true proportion of smartphone users is \(36 \%\) a. What would you expect the shape of the sampling distribution for the sample proportion to be? b. What would be the mean of this sampling distribution? c. What would be the standard deviation of the sampling distribution?

Short Answer

Expert verified
The shape of the sampling distribution for the sample proportion is approximately normal as np and n(1-p) are greater than 5. The mean of this sampling distribution would be 0.36 and the standard deviation would be 0.03416.

Step by step solution

01

Determine the Shape of Sampling Distribution

The shape of the sampling distribution for the sample proportion can be determined by the success/failure condition. If np and n(1-p) are both greater than 5, we can consider the distribution to be approximately normal. Here, n=200 is the sample size and p=0.36 is the proportion of smartphone users. Therefore, np=200*0.36=72 and n(1-p)=200*(1-0.36)=128. Both are greater than 5, so the distribution will be approximately normal.
02

Calculate the Mean of Sampling Distribution

The mean of a sampling distribution (μp̂) is equal to the population proportion (p). Since the population proportion of smartphone users is 0.36, the mean of this sampling distribution would be 0.36.
03

Compute the Standard Deviation of Sampling Distribution

The standard deviation for a sampling distribution (σp̂) can be calculated using the formula: σp̂= sqrt[(p(1-p)/n)]. Substituting the given values into the formula, we get σp̂ = sqrt[(0.36*(1-0.36))/200] = 0.03416

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Key Concepts

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

Success/Failure Condition
Understanding the success/failure condition is key to determining the shape of the sampling distribution for a sample proportion. This condition is part of a set of requirements needed to approximate a binomial distribution with a normal distribution, which is often more practical to work with.

The success/failure condition necessitates that the expected number of successes (np) and the expected number of failures (n(1-p)) in the sample must both be greater than or equal to 5. This rule ensures that the sample size is sufficiently large so that the sampling distribution can be considered symmetric and bell-shaped, resembling the normal distribution that we often rely on for further statistical inferences.

In the given exercise, with a sample size of 200 and a true proportion of smartphone users being 36%, we calculate both np (72) and n(1-p) (128), which are indeed greater than 5. Hence, the condition is satisfied, indicating that the sampling distribution of the sample proportion will likely be normal, allowing us to use z-scores and other normal distribution properties for analysis.
Sample Proportion
The sample proportion, denoted as \( \hat{p} \), represents the fraction of successes in a sample and plays a pivotal role when assessing statistics related to proportions within a population. It's a statistic that estimates a population parameter – the population proportion p.

When a random sample is taken, as in the investment company’s case, the sample proportion is used to estimate the true population proportion of customers using smartphones to access the website. Given a true population proportion of 36%, we can infer, just for this sample, that about 36 out of every 100 customers are expected to access the site via smartphone.

However, due to natural variability, we can expect different samples to yield different sample proportions. This is where the concept of a sampling distribution comes into play – it allows us to understand and quantify the variations among these sample proportions.
Standard Deviation of Sampling Distribution
The standard deviation of the sampling distribution, often denoted as \( \sigma_{\hat{p}} \), measures the variability of the sample proportions from the true population proportion. It's an essential concept because it quantifies how much the sample proportion is likely to fluctuate from sample to sample.

The formula to calculate this standard deviation is \( \sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} \), where p is the population proportion and n is the sample size. A smaller standard deviation indicates that the sample proportions are tightly clustered around the population proportion, whereas a larger standard deviation suggests more spread out sample proportions.

In the exercise, we are given the population proportion (0.36) and the sample size (200), allowing us to compute the standard deviation \( \sigma_{\hat{p}} \) as approximately 0.03416. This value is crucial for constructing confidence intervals or conducting hypothesis tests regarding the population proportion.

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Most popular questions from this chapter

Cars What fraction of cars made in Japan? The computer output below summarizes the results of a random sample of 50 autos. Explain carefully what it tells you. z-Interval for proportion With \(90.00 \%\) confidence, \(0.29938661<\mathrm{P}(\) japan \()<0.46984416\)

Junk mail Direct mail advertisers send solicitations (a.k.a. "junk mail") to thousands of potential customers in the hope that some will buy the company's product. The acceptance rate is usually quite low. Suppose a company wants to test the response to a new flyer, and sends it to 1000 people randomly selected from their mailing list of over 200,000 people. They get orders from 123 of the recipients. a. Create a \(90 \%\) confidence interval for the percentage of people the company contacts who may buy something. b. Explain what this interval means. c. Explain what "90\% confidence" means. d. The company must decide whether to now do a mass mailing. The mailing won't be cost-effective unless it produces at least a \(5 \%\) return. What does your confidence interval suggest? Explain.

Conclusions A catalog sales company promises to deliver orders placed on the Internet within 3 days. Follow-up calls to a few randomly selected customers show that a \(95 \%\) confidence interval for the proportion of all orders that arrive on time is \(88 \% \pm 6 \%\). What does this mean? Are these conclusions correct? Explain. a. Between \(82 \%\) and \(94 \%\) of all orders arrive on time. b. Ninety-five percent of all random samples of customers will show that \(88 \%\) of orders arrive on time. c. Ninety-five percent of all random samples of customers will show that \(82 \%\) to \(94 \%\) of orders arrive on time. d. We are \(95 \%\) sure that between \(82 \%\) and \(94 \%\) of the orders placed by the sampled customers arrived on time. e. On \(95 \%\) of the days, between \(82 \%\) and \(94 \%\) of the orders will arrive on time.

Confidence intervals, again Several factors are involved in the creation of a confidence interval. Among them are the sample size, the level of confidence, and the margin of error. Which statements are true? a. For a given sample size, reducing the margin of error will mean lower confidence. b. For a certain confidence level, you can get a smaller margin of error by selecting a bigger sample. c. For a fixed margin of error, smaller samples will mean lower confidence. d. For a given confidence level, a sample 9 times as large will make a margin of error one third as big.

Conditions For each situation described below, identify the population and the sample, explain what \(p\) and \(\hat{p}\) represent, and tell whether the methods of this chapter can be used to create a confidence interval. a. Police set up an auto checkpoint at which drivers are stopped and their cars inspected for safety problems. They find that 14 of the 134 cars stopped have at least one safety violation. They want to estimate the percentage of all cars that may be unsafe. b. A TV talk show asks viewers to register their opinions on prayer in schools by logging on to a website. Of the 602 people who voted, 488 favored prayer in schools. We want to estimate the level of support among the general public. c. A school is considering requiring students to wear uniforms. The PTA surveys parent opinion by sending a questionnaire home with all 1245 students; 380 surveys are returned, with 228 families in favor of the change. d. A college admits 1632 freshmen one year, and four years later, 1388 of them graduate on time. The college wants to estimate the percentage of all their freshman enrollees who graduate on time.

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