In an attempt to increase business on Monday nights, a restaurant offers a free dessert with every dinner order. Before the offer, the mean number of dinner customers on Monday was 150. Following are the numbers of diners on a random sample of 12 days while the offer was in effect. Can you conclude that the mean number of diners decreased while the free dessert offer was in effect? Use the a=0.01 level of significance and the p-value method with the TI-84 Plus calculator.

Question

Compute the value of the test statistic

Answer 1

To complete the hypothesis test regarding whether the mean number of diners decreased with the free dessert offer, we'll follow these steps:

1. State the Hypotheses:

   • Null Hypothesis (\(H_0\)): \(\mu = 150\) (the mean number of diners did not decrease)
   • Alternative Hypothesis (\(H_a\)): \(\mu < 150\) (the mean number of diners decreased)

2. Collect the Sample Data:

   • You mentioned a random sample of 12 days, but you didn't provide the specific numbers of diners for those days. To proceed, you'll need to provide that data.

3. Calculate the Sample Mean and Standard Deviation:

   • Once you have the data, compute the sample mean (\(\bar{x}\)) and sample standard deviation (\(s\)).

4. Calculate the Test Statistic:

   • The test statistic for a t-test is calculated using the formula: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \] Where:
   • \(\bar{x}\) is the sample mean
   • \(\mu_0\) is the population mean under the null hypothesis (150)
   • \(s\) is the sample standard deviation
   • \(n\) is the sample size (12)

Please provide the sample data to proceed with the calculations, or if you have the computed sample mean and standard deviation, please share those values. This way, I can help you compute the test statistic.

Answer 2

the numbers are 170, 133, 150, 111, 171, 103, 101, 110, 133, 179, 151, 112

Answer 3

Great! Now that we have the data, we can follow the steps to calculate the test statistic and conduct the hypothesis test.

Given Data:

The sample of diner counts is: \[ 170, 133, 150, 111, 171, 103, 101, 110, 133, 179, 151, 112 \]

Step 1: Calculate the Sample Mean (\(\bar{x}\)) and Sample Standard Deviation (s)

1. Calculate the Sample Mean (\(\bar{x}\)): \[ \bar{x} = \frac{\sum x_i}{n} \]

   Where:

   • \(n = 12\) (the number of days)
   • \(x_i\) are the individual counts.

   \[ \bar{x} = \frac{170 + 133 + 150 + 111 + 171 + 103 + 101 + 110 + 133 + 179 + 151 + 112}{12} \]

   \[ \bar{x} = \frac{1,524}{12} \approx 127.00 \]

2. Calculate the Sample Standard Deviation (\(s\)):

   • First, find the variance: \[ s^2 = \frac{\sum (x_i - \bar{x})^2}{n - 1} \]

   First, we need to calculate \((x_i - \bar{x})^2\) for each value:

   • (170 - 127)^2 = 1849
   • (133 - 127)^2 = 36
   • (150 - 127)^2 = 529
   • (111 - 127)^2 = 256
   • (171 - 127)^2 = 1936
   • (103 - 127)^2 = 576
   • (101 - 127)^2 = 676
   • (110 - 127)^2 = 289
   • (133 - 127)^2 = 36
   • (179 - 127)^2 = 2704
   • (151 - 127)^2 = 576
   • (112 - 127)^2 = 225

   Now, summing those: \[ \sum (x_i - \bar{x})^2 = 1849 + 36 + 529 + 256 + 1936 + 576 + 676 + 289 + 36 + 2704 + 576 + 225 = 11511 \]

   Plugging into the variance formula: \[ s^2 = \frac{11511}{11} \approx 1046.4545 \] \[ s \approx \sqrt{1046.4545} \approx 32.35 \]

Step 2: Calculate the Test Statistic

Using the formula: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \] Where:

• \(\mu_0 = 150\)
• \(\bar{x} \approx 127.00\)
• \(s \approx 32.35\)
• \(n = 12\)

Calculate the standard error (SE): \[ SE = \frac{s}{\sqrt{n}} = \frac{32.35}{\sqrt{12}} \approx \frac{32.35}{3.464} \approx 9.35 \]

Now calculate the test statistic \(t\): \[ t = \frac{127.00 - 150}{9.35} = \frac{-23}{9.35} \approx -2.46 \]

Final Result

The value of the test statistic \(t\) is approximately -2.46.

The Next Steps

If required, you can then calculate the p-value corresponding to this t-test statistic using a calculator such as the TI-84, or based on the critical t-value for a one-tailed test with \(df = n - 1 = 11\) at the \(\alpha = 0.01\) level of significance, in which you'd check if the p-value < \(\alpha\) to determine whether to reject the null hypothesis or not.