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Chapter 10: Problem 19

A Washington, D.C., "think tank" announces the typical teenager sent 67 text messages per day in 2017 . To update that estimate, you phone a sample of 12 teenagers and ask them how many text messages they sent the previous day. Their responses were: \(\begin{array}{llllllllll}51 & 175 & 47 & 49 & 44 & 54 & 145 & 203 & 21 & 59 & 42 & 100\end{array}\) At the .05 level, can you conclude that the mean number is greater than \(67 ?\) Estimate the \(p\) -value and describe what it tells you.

Short Answer

Expert verified
The test does not provide enough evidence to conclude that the mean is greater than 67.

Step by step solution

01

Calculate the Sample Mean

First, we need to calculate the sample mean of the given number of text messages. The formula for the sample mean \( \bar{x} \) is:\[ \bar{x} = \frac{\sum x_i}{n} \]where \( x_i \) are the individual sample values, and \( n \) is the sample size.For the given data: \[ \sum x_i = 51 + 175 + 47 + 49 + 44 + 54 + 145 + 203 + 21 + 59 + 42 + 100 = 990 \]\( n = 12 \).Thus, the sample mean \( \bar{x} = \frac{990}{12} = 82.5 \).
02

Calculate the Sample Standard Deviation

Next, calculate the sample standard deviation \( s \) to assess the variability in the data. The formula for the sample standard deviation is:\[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}} \]Calculate each squared difference and then the sum:- \[ (51 - 82.5)^2 = 992.25, \, (175 - 82.5)^2 = 8585.25, \, \ldots, \, (100 - 82.5)^2 = 306.25 \]- Sum of squared differences: \[ \sum (x_i - \bar{x})^2 = 9973.5 \]Calculate the standard deviation:\[ s = \sqrt{\frac{9973.5}{11}} \approx 30.03 \]
03

Set Up the Hypothesis

Define the null hypothesis \( H_0 \) and the alternative hypothesis \( H_a \):- \( H_0: \mu = 67 \), the mean number of text messages is 67.- \( H_a: \mu > 67 \), the mean number of text messages is greater than 67.We will test this using a one-tailed t-test.
04

Calculate the t-Statistic

Calculate the t-statistic using the formula:\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]Where \( \mu_0 = 67 \).Substitute the values:\[ t = \frac{82.5 - 67}{30.03 / \sqrt{12}} \approx \frac{15.5}{8.67} \approx 1.79 \]
05

Determine the Critical Value and Compare

Using a t-distribution table with \( n-1 = 11 \) degrees of freedom and \( \alpha = 0.05 \) for a one-tailed test, the critical value \( t_c \approx 1.796 \).Since our calculated t-value (1.79) is less than \( t_c \), we do not reject the null hypothesis \( H_0 \).
06

Estimate the p-value

The p-value reflects the probability of obtaining a test statistic at least as extreme as the one computed. For \( t = 1.79 \) and 11 degrees of freedom, the p-value is slightly greater than 0.05. Since \( p > 0.05 \), we fail to reject the null hypothesis.
07

Conclusion: Decision and Interpretation

Based on the p-value and comparison with the critical value, there is insufficient evidence to conclude that the mean number of text messages is greater than 67 at the 0.05 significance level.

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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 that represents the average value of a sample. It provides a single value summary of a set of numbers and is calculated by adding up all the individual observations and then dividing by the total number of observations in the sample.
  • To calculate the sample mean, use the formula \( \bar{x} = \frac{\sum x_i}{n} \), where \( x_i \) are the data points and \( n \) is the sample size.
  • In this problem, the sample size \( n \) is 12, as there are 12 teenagers surveyed about their text messages.
  • The total number of text messages sent as given by the teenagers is 990. Dividing this sum by the number of teenagers gives us a sample mean \( \bar{x} = 82.5 \).
The sample mean of 82.5 indicates that, on average, the teenagers in this sample sent approximately 82.5 text messages the previous day, which helps assess whether this differs significantly from the presumed population mean of 67 messages.
Sample Standard Deviation
Sample standard deviation is a measure of the amount of variation or dispersion in a set of values. It assesses whether the data points are spread out over a wide range or clustered closely around the sample mean. A larger standard deviation indicates a wide spread of numbers, while a smaller one suggests closer clustering.
  • To compute the sample standard deviation \( s \), use the formula \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}} \]. Here, \( x_i \) are the individual data points, \( \bar{x} \) is the sample mean, and \( n-1 \) represents the degrees of freedom.
  • In our example, the deviations of individual text message counts from the mean (82.5) are squared, summed (9973.5), and then divided by the degrees of freedom (11).
  • The result \( s \approx 30.03 \) gives us an idea of how varied the teenagers' text message counts are around the mean of 82.5.
The sample standard deviation of 30.03 suggests moderate variability in the teenagers' text message sending behavior. This variability is essential for accurately conducting the subsequent hypothesis test.
T-Test
The t-test is a statistical method used to determine if there is a significant difference between the means of two groups or if a sample mean significantly deviates from a known value. It is particularly useful when dealing with small sample sizes and when the population standard deviation is unknown.
  • The t-test compares the calculated t-statistic against a critical value to decide whether to reject the null hypothesis.
  • For this exercise, the t-test checks if the mean number of text messages sent, as calculated from our sample, is greater than the expected mean of 67 messages.
  • The formula for the t-statistic is \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \], where \( \bar{x} \) is the sample mean, \( \mu_0 \) is the hypothesized population mean, \( s \) is the sample standard deviation, and \( n \) is the sample size.
  • Using our data, we calculate \( t \approx 1.79 \).
The calculated t-value of 1.79 needs to be compared with the critical t-value from a t-distribution table to determine the significance of the results.
P-Value Estimation
The p-value in hypothesis testing is the probability of observing a test statistic as extreme as the one obtained, assuming the null hypothesis is true. In simpler terms, it tells you how likely it is to get your current results, given that the null hypothesis is correct.
  • The smaller the p-value, the stronger the evidence against the null hypothesis. If the p-value is less than the significance level (often 0.05), the null hypothesis is rejected.
  • In this problem, after calculating the t-statistic (1.79), the p-value is estimated by comparing it against the t-distribution with 11 degrees of freedom.
  • The p-value here is slightly greater than 0.05, indicating insufficient evidence to reject the null hypothesis that the mean number of messages is 67.
Since the p-value is not low enough to discard the null hypothesis, we conclude with this sample that there's not enough evidence to say the average number of messages sent by teenagers is greater than 67.

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Most popular questions from this chapter

The amount of water consumed each day by a healthy adult follows a normal distribution with a mean of 1.4 liters. A health campaign promotes the consumption of at least 2.0 liters per day. A sample of 10 adults after the campaign shows the following consumption in liters: $$\begin{array}{llllllll}1.5 & 1.6 & 1.5 & 1.4 & 1.9 & 1.4 & 1.3 & 1.9 & 1.8 & 1.7\end{array}$$ At the .01 significance level, can we conclude that water consumption has increased? Calculate and interpret the \(p\) -value.

For a recent year, the mean fare to fly from Charlotte, North Carolina, to Chicago, Illinois, on a discount ticket was \(\$ 267 .\) A random sample of 13 round-trip discount fares on this route last month shows: \(\begin{array}{llllllllll}\$ 321 & \$ 286 & \$ 290 & \$ 330 & \$ 310 & \$ 250 & \$ 270 & \$ 280 & \$ 299 & \$ 265 & \$ 291 & \$ 275 & \$ 281\end{array}\) At the .01 significance level, can we conclude that the mean fare has increased? What is the \(p\) -value?

Given the following hypotheses: $$\begin{array}{l}H_{0}: \mu=100 \\\H_{1}: \mu \neq 100\end{array}$$ A random sample of six resulted in the following values: \(118,105,112,119,105,\) and 111 . Assume a normal population. Using the .05 significance level, can we conclude the mean is different from \(100 ?\) a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis? d. Estimate the \(p\) -value.

A recent article in The Wall Street Journal reported that the home equity loan rate is now less than \(4 \% .\) A sample of eight small banks in the Midwest revealed the following home equity loan rates (in percent): \(\begin{array}{lllllll}3.6 & 4.1 & 5.3 & 3.6 & 4.9 & 4.6 & 5.0 & 4.4\end{array}\) At the .01 significance level, can we conclude that the home equity loan rate for small banks is less than \(4 \%\) ? Estimate the \(p\) -value.

Answer the questions: (a) Is this a one- or two-tailed test? (b) What is the decision rule? (c) What is the value of the test statistic? (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. A sample of 36 observations is selected from a normal population. The sample mean is \(12,\) and the population standard deviation is \(3 .\) Conduct the following test of hypothesis using the .01 significance level. $$ \begin{array}{l} H_{0}: \mu \leq 10 \\ H_{1}: \mu>10 \end{array} $$
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