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CCN and ActMedia provided a television channel targeted to individuals waiting in supermarket checkout lines. The channel showed news, short features, and advertisements. The length of the program was based on the assumption that the population mean time a shopper stands in a supermarket checkout line is 8 minutes. A sample of actual waiting times will be used to test this assumption and determine whether actual mean waiting time differs from this standard. a. Formulate the hypotheses for this application. b. A sample of 120 shoppers showed a sample mean waiting time of 8.4 minutes. Assume a population standard deviation of \(\sigma=3.2\) minutes. What is the \(p\) -value? c. \(\quad\) At \(a=.05,\) what is your conclusion? d. Compute a \(95 \%\) confidence interval for the population mean. Does it support your conclusion?

Short Answer

Expert verified
The mean waiting time does not significantly differ from 8 minutes. Confidence interval: (7.827, 8.973).

Step by step solution

01

Formulate Hypotheses

First, we need to set up our null and alternative hypotheses. The null hypothesis, \(H_0\), is that the population mean \(\mu\) is equal to 8 minutes. The alternative hypothesis, \(H_a\), is that the population mean \(\mu\) is not equal to 8 minutes (two-tailed test). Thus, the hypotheses are:\[H_0: \mu = 8 \H_a: \mu eq 8\]
02

Calculate the Test Statistic

We will use the z-test since the population standard deviation is known. The formula for the z-test statistic is:\[z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\]Substituting the given values: \(\bar{x} = 8.4\), \(\mu = 8\), \(\sigma = 3.2\), and \(n = 120\), we have:\[z = \frac{8.4 - 8}{\frac{3.2}{\sqrt{120}}} = \frac{0.4}{0.2929} \approx 1.366\]
03

Find the p-value

Using the z-value calculated (\(z \approx 1.366\)), we find the p-value for a two-tailed test. Consulting standard normal distribution tables or using software, we find that:- The p-value for \(z = 1.366\) is approximately 0.172.Since this is a two-tailed test, we account for both tails, giving us the p-value \(p \approx 0.172\).
04

Draw a Conclusion (Significance Level \(\alpha = 0.05\))

With \(p = 0.172\), compare it to the significance level \(\alpha = 0.05\).Since \(p > \alpha\), we do not reject the null hypothesis. There is not enough statistical evidence to say the mean waiting time differs from 8 minutes.
05

Calculate the Confidence Interval

The formula for a 95% confidence interval for the mean is given by:\[\bar{x} \pm z_{\frac{\alpha}{2}} \times \frac{\sigma}{\sqrt{n}}\]Where \(z_{\frac{\alpha}{2}} = 1.96\) for a 95% confidence level. Substituting the values, we get:\[8.4 \pm 1.96 \times \frac{3.2}{\sqrt{120}} \ 8.4 \pm 0.573\]Thus, the confidence interval is approximately (7.827, 8.973).
06

Analyze the Confidence Interval

The 95% confidence interval (7.827, 8.973) includes the value 8 minutes. This supports our earlier conclusion that there is no significant difference from the assumed mean of 8 minutes.

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Key Concepts

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

Sample Mean
The sample mean is a fundamental concept in statistics, representing the average of a set of observations. In this exercise, the sample mean is used to estimate the average time shoppers spend at checkout lines. Calculating the sample mean involves summing up all observed waiting times and then dividing by the number of observations, or \[ \bar{x} = \frac{\sum_{i=1}^{n}x_i}{n} \]Where:
  • \(\bar{x}\) is the sample mean,
  • \(x_i\) are individual sample observations,
  • \(n\) is the sample size.
In our exercise, 120 shoppers were sampled, and the average waiting time calculated was 8.4 minutes. It serves as an estimate of the true population mean, which helps us understand the central tendency of our sampled data. This value becomes a key player when performing statistical tests, like hypothesis testing, as it gives us an initial insight into whether or not the waiting times align with expected values.
Confidence Interval
Confidence intervals provide a range of values, derived from the sample mean, within which we expect the population mean to fall. In essence, it's like stating that we are "confident" the true mean might lie in this interval if we were to take many samples. This is a fantastic way to account for sampling variability.To compute a confidence interval for a population mean, we use the formula:\[ \bar{x} \pm z_{\frac{\alpha}{2}} \times \frac{\sigma}{\sqrt{n}} \]Where:
  • \(\bar{x}\) is the sample mean,
  • \(z_{\frac{\alpha}{2}}\) is the critical value from the normal distribution for our confidence level (e.g., 1.96 for 95%),
  • \(\sigma\) is the population standard deviation,
  • \(n\) is the sample size.
The confidence interval calculated in the exercise was (7.827, 8.973), meaning we are 95% sure that the actual mean waiting time is between these two values. It's crucial to note that if a confidence interval includes the hypothesized mean (8 minutes in this case), it supports the notion that there's no significant difference from the expected value.
p-value
The p-value is a critical concept in hypothesis testing, providing a measure of the probability that the observed results (or more extreme) could occur under the null hypothesis. A smaller p-value indicates that, assuming the null hypothesis is true, the observed data is less likely.In hypothesis testing, we compare the p-value to a predefined significance level \(\alpha\) (commonly 0.05). If the p-value is less than \(\alpha\), it suggests that the data provides enough evidence to reject the null hypothesis.For the exercise, the calculated p-value was approximately 0.172. Given \(\alpha = 0.05\), we find \(p > \alpha\), meaning we do not reject the null hypothesis. This suggests that there isn't strong enough evidence to claim that the mean waiting time significantly differs from the assumed value (8 minutes). Thus, the p-value plays a key role in decision-making processes for statistical hypotheses.

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Most popular questions from this chapter

The label on a 3 -quart container of orange juice states that the orange juice contains an average of 1 gram of fat or less. Answer the following questions for a hypothesis test that could be used to test the claim on the label. a. Develop the appropriate null and alternative hypotheses. b. What is the Type I error in this situation? What are the consequences of making this error? c. What is the Type II error in this situation? What are the consequences of making this error?

AOL Time Warner Inc.'s CNN has been the longtime ratings leader of cable television news. Nielsen Media Research indicated that the mean CNN viewing audience was 600,000 viewers per day during 2002 (The Wall Street Journal, March 10,2003 ). Assume that for a sample of 40 days during the first half of \(2003,\) the daily audience was 612,000 viewers with a sample standard deviation of 65,000 viewers. a. What are the hypotheses if CNN management would like information on any change in the CNN viewing audience? b. What is the \(p\) -value? c. Select your own level of significance. What is your conclusion? d. What recommendation would you make to CNN management in this application?

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Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu \geq 80 \\ H_{\mathrm{a}}: \mu<80 \end{array} \\] A sample of 100 is used and the population standard deviation is \(12 .\) Compute the \(p\) -value and state your conclusion for each of the following sample results. Use \(\alpha=.01\) a. \(\quad \bar{x}=78.5\) b. \(\bar{x}=77\) c. \(\bar{x}=75.5\) d. \(\bar{x}=81\)

In \(2008,46 \%\) of business owners gave a holiday gift to their employees. A 2009 survey of business owners indicated that \(35 \%\) planned to provide a holiday gift to their employees (Radio WEZV, Myrtle Beach, SC, November 11,2009 ). Suppose the survey results are based on a sample of 60 business owners. a. How many business owners in the survey planned to provide a holiday gift to their employees in \(2009 ?\) b. Suppose the business owners in the sample did as they planned. Compute the \(p\) -value for a hypothesis test that can be used to determine if the proportion of business owners providing holiday gifts had decreased from the 2008 level. c. Using a .05 level of significance, would you conclude that the proportion of business owners providing gifts decreased? What is the smallest level of significance for which you could draw such a conclusion?

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