91Ó°ÊÓ

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. \(\quad\) A sample of 120 shoppers showed a sample mean waiting time of 8.5 minutes. Assume a population standard deviation \(\sigma=3.2\) minutes. What is the \(p\) -value? c. \(\quad\) At \(\alpha=.05,\) what is your conclusion? d. Compute a \(95 \%\) confidence interval for the population mean. Does it support your conclusion?

Step by step solution

01

Formulate Hypotheses

We need to formulate the null and alternative hypotheses to test whether the actual mean waiting time differs from 8 minutes. - Null Hypothesis \( H_0: \mu = 8 \) minutes (The mean waiting time is 8 minutes.) - Alternative Hypothesis \( H_a: \mu eq 8 \) minutes (The mean waiting time is not 8 minutes.)

02

Calculate Test Statistic

To find the p-value, calculate the test statistic using the sample mean, population mean, population standard deviation, and sample size. The formula is:\[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]where \( \bar{x} = 8.5 \), \( \mu = 8 \), \( \sigma = 3.2 \), and \( n = 120 \).Substituting the values:\[ z = \frac{8.5 - 8}{3.2 / \sqrt{120}} = \frac{0.5}{3.2 / 10.95} = 1.71 \]

03

Find the P-value

Using a standard normal distribution table or calculator, find the p-value associated with a two-tailed test for \( z = 1.71 \). This p-value is roughly 0.088.

04

Make a Decision

Compare the p-value to the significance level \( \alpha = 0.05 \). Since the p-value (0.088) is greater than \( \alpha \), we do not reject the null hypothesis. Conclusion: There is not enough statistical evidence to conclude that the mean waiting time differs from 8 minutes.

05

Compute Confidence Interval

Calculate the 95% confidence interval for the mean using the formula:\[ \bar{x} \pm z^* \left( \frac{\sigma}{\sqrt{n}} \right) \]Where \( \bar{x} = 8.5 \), \( z^* = 1.96 \) for 95% confidence, \( \sigma = 3.2 \), and \( n = 120 \).The interval is:\[ 8.5 \pm 1.96 \left( \frac{3.2}{10.95} \right) \approx 8.5 \pm 0.572 \]So, the confidence interval is \([7.928, 9.072]\).

06

Interpret Confidence Interval

The 95% confidence interval \([7.928, 9.072]\) contains the hypothesized mean of 8 minutes. This supports our earlier conclusion that there is not enough evidence to suggest the mean waiting time differs significantly from 8 minutes.

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Key Concepts

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

Confidence Interval

A confidence interval is a range of values that estimates a population parameter, such as a mean, with a certain level of confidence. In the context of the supermarket example, we want a range where we believe the true average waiting time falls.
For a 95% confidence interval, we are saying that if we took 100 different samples and built confidence intervals for each one, approximately 95 of them would contain the true population mean.
In our problem, the sample mean was 8.5 minutes, and using the formula \[ \bar{x} \pm z^* \left( \frac{\sigma}{\sqrt{n}} \right) \] where \( \bar{x} = 8.5 \), \( z^* = 1.96 \), and \( \sigma = 3.2 \), the confidence interval calculated is \[ [7.928, 9.072] \].
This range includes the hypothesized mean of 8 minutes, suggesting that our assumption is not necessarily wrong.

P-value

The p-value helps us determine the strength of our results in hypothesis testing. It represents the probability that the observed data (or something more extreme) would occur if the null hypothesis were true.
In simpler terms, a small p-value implies that the observed data would be very rare under the null hypothesis, suggesting that we might have enough evidence to reject the null hypothesis.
For our exercise, the p-value was approximately 0.088, derived from a test statistic \( z = 1.71 \).
Because this value is greater than our significance level of 0.05, we do not reject the null hypothesis. Meaning, the data does not provide strong enough evidence to suggest a difference from an 8-minute average.

Population Mean

The population mean is a measure representing the average value of a group. It's a crucial element in hypothesis testing because it serves as a hypothetical benchmark.
In our example, the population mean was assumed to be 8 minutes, representing the average waiting time in checkout lines.
When we talk about checking hypotheses, we're often concerned with whether our sample (like the 8.5-minute sample mean) tells us something new about the entire population.
If our sample suggests the population mean might be different than 8 minutes, we test that hypothesis using statistical tools to decide if the difference is significant.

Z-test

A z-test is a statistical test used to determine if there is a significant difference between the sample mean and a known population mean. It is ideal when dealing with large sample sizes and known population standard deviations.
The formula used is \[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \], where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
In our scenario, a z-test was conducted, finding a test statistic of 1.71. A corresponding p-value helped determine whether we have enough evidence to reject the null hypothesis.
The z-test underpins much of hypothesis testing by showing, in standard deviation units, how far our sample mean is from the population mean, thus guiding decisions about what is statistically significant.