91Ó°ÊÓ

The past records of a supermarket show that its customers spend an average of \(\$ 95\) per visit at this store. Recently the management of the store initiated a promotional campaign according to which each customer receives points based on the total money spent at the store, and these points can be used to buy products at the store. The management expects that as a result of this campaign, the customers should be encouraged to spend more money at the store. To check whether this is true, the manager of the store took a sample of 14 customers who visited the store. The following data give the money (in dollars) spent by these customers at this supermarket during their visits. \(\begin{array}{rrrrrrr}109.15 & 136.01 & 107.02 & 116.15 & 101.53 & 109.29 & 110.79 \\ 94.83 & 100.91 & 97.94 & 104.30 & 83.54 & 67.59 & 120.44\end{array}\) Assume that the money spent by all customers at this supermarket has a normal distribution. Using a \(5 \%\) significance level, can you conclude that the mean amount of money spent by all customers at this supermarket after the campaign was started is more than \(\$ 95\) ? (Hint: First calculate the sample mean and the sample standard deviation for these data using the formulas learned in Sections \(3.1 .1\) and \(3.2 .2\) of Chapter 3 . Then make the test of hypothesis about \(\mu .\) )

Step by step solution

01

Calculate Sample Mean

First, sum up all the given data values and divide it by the total number of values (14 in this case) to calculate the sample mean. Let's denote this value as \(\overline{X}\).

02

Calculate Sample Standard Deviation

Subtract each data value from the mean \(\overline{X}\), square the results, and then sum them up. Divide this sum by the total number of values minus 1 (14-1), and then take the square root of the result to get the sample standard deviation \(s\).

03

Hypothesis Formulation

Formulate the null hypothesis as \(H_0: \mu = \$95\) and the alternative hypothesis as \(H_A: \mu > \$95\). This is because we want to check if the mean amount of money spent is more than \$95.

04

Test Statistic Calculation

Use the formula for the test statistic for a sample mean, which is \(Z = (\overline{X} - \mu) / \frac{s}{\sqrt{n}}\), where \(\overline{X}\) is the sample mean, \(\mu\) is the null hypothesis mean (in this case, \$95), \(s\) is the sample standard deviation, and \(n\) is the number of data values (in this case, 14). Calculate this statistic.

05

Find the Critical Value and Take Decision

Find the critical value \(Z_{critical}\) from the standard normal distribution table for the significance level of 0.05. This is the value where to the left is 95% area under the curve and to the right is 5% (or 0.05). If the computed test statistic \(Z\) is greater than the critical value \(Z_{critical}\), then reject the null hypothesis.

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Key Concepts

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

Normal Distribution

The normal distribution is a foundational concept in statistics, often described as a bell-shaped curve. It is symmetric about the mean, which means it looks the same to the left and right of the center point. Imagine drawing a line through the middle of the curve; both sides mirror each other. Here are some essential characteristics of a normal distribution:

• The mean, median, and mode of a normal distribution are all equal and located at the center.
• It is determined by two main parameters: the mean (average) and the standard deviation (how spread out the values are).
• The total area under the curve equals 1, representing 100% of the data.
• Approximately 68% of data fall within one standard deviation of the mean, about 95% within two, and around 99.7% within three, known as the empirical rule or 68-95-99.7 rule.

Understanding the normal distribution allows statisticians to make predictions about data and is crucial when performing hypothesis testing, like in the supermarket scenario.

Sample Mean

The sample mean is an estimate of the population mean based on a subset, or sample, from the entire population. In the context of the supermarket example, the manager uses the sample mean to estimate the average amount spent by all customers. Here’s how it’s calculated:

• Add up all the data values in the sample. For our supermarket case, these are the amounts spent by the 14 customers.
• Divide this sum by the number of data points, which is 14 here.

This gives you the sample mean, denoted as \(\overline{X}\). The sample mean provides a snapshot of the central tendency of the data, reflecting how much customers approximately spend during a visit after the campaign.

Standard Deviation

Standard deviation measures the amount of variation or dispersion in a set of values. In simpler terms, it tells us how much the values in a dataset deviate from the mean:

• A low standard deviation means that the data points tend to be close to the mean.
• A high standard deviation indicates that the data are spread out over a wider range of values.

To calculate the sample standard deviation:

• Find the difference between each data value and the sample mean (\(\overline{X}\)).
• Square these differences to make them positive and then sum them all up.
• Divide this sum by the number of data points minus one (\(n-1\), where \(n\) is 14).
• Finally, take the square root of the result to obtain the standard deviation \(s\).

In the supermarket example, the standard deviation helps to understand the variability in the spending behavior of customers.

Significance Level

The significance level, often denoted as \(\alpha\), is the probability of rejecting the null hypothesis when it is true. It's a threshold set by the researcher before the study to determine how much evidence you need to reject the null hypothesis:

• A common choice is 5% (0.05), which means there is a 5% risk of concluding that a difference exists when there is none.
• This level indicates how extreme the data has to be before we can reject the null hypothesis.

In our supermarket scenario, the significance level is set at 0.05, guiding the decision-making process on whether the points campaign significantly increased customer spending. It determines the critical value needed from the standard normal distribution table to compare with the calculated test statistic. Hence, it plays a pivotal role in the hypothesis testing process, influencing the final conclusion.