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Chapter 10: Problem 78

An economist was interested in studying the impact of the recession on dining out, including drive-thru meals at fast food restaurants. A random sample of forty-eight families of four with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week indicated that they reduced their spending on dining out by an average of \(\$ 31.47\) per week, with a sample standard deviation of \(\$ 10.95\). Another random sample of 42 families of five with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week reduced their spending on dining out by an average \(\$ 35.28\) per week, with a sample standard deviation of \(\$ 12.37\). (Note that the two groups of families are differentiated by the number of family members.) Assume that the distributions of reductions in weekly dining-out spendings for the two groups have the same population standard deviation. a. Construct a \(90 \%\) confidence interval for the difference in the mean weekly reduction in diningout spending levels for the two populations. b. Using the \(5 \%\) significance level, can you conclude that the average weekly spending reduction for all families of four with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week is less than the average weekly spending reduction for all families of five with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week?

Short Answer

Expert verified
The exact confidence interval and the result of the hypothesis test would depend upon the calculated values, but the above steps detail how to arrive at those outcomes. The short answer would be in the form of: 'The 90% confidence interval for the difference in means is from _ to _. Based on the hypothesis test, we can (not) conclude that the average weekly spending reduction for families of four is less than that of families of five.'

Step by step solution

01

Identify the known and unknown variables

In the first group, the number of families, \(n_1\) is 48, mean spending reduction, \(\bar{x}_1\) is $31.47 and standard deviation, \(s_1\) is $10.95. In the second group, \(n_2\) is 42, \(\bar{x}_2\) is $35.28 and \(s_2\) is $12.37. We are to find the difference of means confidence interval and perform a hypothesis testing.
02

Calculate the standard error

Since it's assumed that the population standard deviations are equal, we will calculate the pooled standard deviation, \(s_p =\sqrt{((n_1-1)s_1^2 + (n_2-1)s_2^2) /(n_1+n_2-2)} \) and use it to find the standard error, \( SE = s_p \sqrt{1/n_1 + 1/n_2} \).
03

Calculate the confidence interval for the difference in means

Using the standard error from Step 2, now calculate the 90% confidence interval for the difference in means. \( (\bar{x}_1-\bar{x}_2) \pm t_{0.05} \times SE \) where \( t_{0.05} \) is the t-value for a 5% significance level with \(n_1 + n_2 - 2\) degrees of freedom.
04

Perform the hypothesis test

Now, set up the null, \(H_0: \mu_1 = \mu_2 \), and alternative, \(H_a: \mu_1 < \mu_2 \), hypotheses. Calculate the t-statistic, \(t = (\bar{x}_1 - \bar{x}_2)/SE \). If \( t < -t_{0.05} \), we reject \(H_0 \) and conclude that the average weekly spending reduction for families of four is less than that of families of five.

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Key Concepts

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

Hypothesis Testing
In hypothesis testing, we assess if a hypothesis about a population parameter is supported by sample data. Here, we investigate whether families of four reduced their weekly dining spending differently from families of five. We have two hypotheses:
  • Null hypothesis (\(H_0\)): The spending reductions for both groups are equal, \( \mu_1 = \mu_2\).
  • Alternative hypothesis (\(H_a\)): The reduction for families of four is less than that for families of five, \( \mu_1 < \mu_2\).
This involves determining the probability of observing the sample results if the null hypothesis is true. Statistical significance, often tested with a 5% significance level, helps us decide whether to reject the null hypothesis. In this context, if our calculated t-value is less than the critical value, we reject \(H_0\), accepting the notion that families of four spend less on dining out.
Standard Error
Standard error reflects the variability of a sample mean estimate around the population mean. It gives us an idea of how much our sample mean might differ from the true population mean. It is a crucial component in calculating confidence intervals and in hypothesis testing.
  • For individual groups, the standard error (\(SE\)) derives from their standard deviations and sample sizes. However, in our study, due to the assumption of equal population variances, we utilize a pooled standard deviation to compute a combined standard error.
  • The formula used is \(SE = s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}\), where \(s_p\) is the pooled standard deviation. This equation helps in estimating how the sample mean difference provides insights into the actual population mean differences.
Pooled Standard Deviation
The pooled standard deviation (\(s_p\)) is pivotal when we assume multiple groups have an equivalent population variance. It offers a combined measure of variance for the samples being analyzed together.
  • The calculation involves averaging the variances weighted by the degrees of freedom for each sample group. The formula is:\[s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}}\]
  • In our scenario, \(s_1\) and \(s_2\) are the sample standard deviations of the two groups. Using these values and the sample sizes, the pooled standard deviation consolidates this data into one metric.
  • By using \(s_p\), we ensure our standard error and confidence intervals are founded on the shared assumptions of variance, leading to more accurate hypothesis testing outcomes.
T-Distribution
The t-distribution is crucial when estimating the mean differences from sample data, especially when dealing with smaller sample sizes or unknown population variances. It resembles the normal distribution but with heavier tails, meaning it is more adept at capturing variability in data.
  • It is often used in hypothesis testing and confidence intervals, allowing us to draw conclusions from small samples.
  • The t-distribution is defined by the degrees of freedom, \(df = n_1 + n_2 - 2\) in our case. As the degrees of freedom increase (with larger sample sizes), it approaches a normal distribution.
  • In constructing a confidence interval or determining a critical t-value for hypothesis testing, we refer to the t-distribution, ensuring we account for variability and sample size influences.

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Most popular questions from this chapter

The owner of a mosquito-infested fishing camp in Alaska wants to test the effectiveness of two rival brands of mosquito repellents, \(\mathrm{X}\) and \(\mathrm{Y}\). During the first month of the season, eight people are chosen at random from those guests who agree to take part in the experiment. For each of these guests, Brand \(\mathrm{X}\) is randomly applied to one arm and Brand \(\mathrm{Y}\) is applied to the other arm. These guests fish for 4 hours, then the owner counts the number of bites on each arm. The table below shows the number of bites on the arm with Brand \(X\) and those on the arm with Brand \(Y\) for each guest. $$ \begin{array}{l|rrrrrrrr} \hline \text { Guest } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } & \text { F } & \text { G } & \text { H } \\ \hline \text { Brand X } & 12 & 23 & 18 & 36 & 8 & 27 & 22 & 32 \\ \hline \text { Brand Y } & 9 & 20 & 21 & 27 & 6 & 18 & 15 & 25 \\ \hline \end{array} $$ a. Construct a \(95 \%\) confidence interval for the mean \(\mu_{d}\) of population paired differences, where a paired difference is defined as the number of bites on the arm with Brand \(X\) minus the number of bites on the arm with Brand \(\mathrm{Y}\). b. Test at the \(5 \%\) significance level whether the mean number of bites on the \(\operatorname{arm}\) with Brand \(\mathrm{X}\) and the mean number of bites on the arm with Brand \(\mathrm{Y}\) are different for all such guests.

Explain when you would use the paired-samples procedure to make confidence intervals and test hypotheses.

The manager of a factory has devised a detailed plan for evacuating the building as quickly as possible in the event of a fire or other emergency. An industrial psychologist believes that workers actually leave the factory faster at closing time without following any system. The company holds fire drills periodically in which a bell sounds and workers leave the building according to the system. The evacuation time for each drill is recorded. For comparison, the psychologist also records the evacuation time when the bell sounds for closing time each day. A random sample of 36 fire drills showed a mean evacuation time of \(5.1\) minutes with a standard deviation of \(1.1\) minutes. A random sample of 37 days at closing time showed a mean evacuation time of \(4.2\) minutes with a standard deviation of \(1.0\) minute. a. Construct a \(99 \%\) confidence interval for the difference between the two population means. b. Test at the \(5 \%\) significance level whether the mean evacuation time is smaller at closing time than during fire drills.

The following information was obtained from two independent samples selected from two normally distributed populations with unknown but equal standard deviations. $$ \begin{array}{lllllllllllll} \text { Sample 1: } & 2.18 & 2.23 & 1.96 & 2.24 & 2.72 & 1.87 & 2.68 & 2.15 & 2.49 & 2.05 & & \\ \text { Sample 2: } & 1.82 & 1.26 & 2.00 & 1.89 & 1.73 & 2.03 & 1.43 & 2.05 & 1.54 & 2.50 & 1.99 & 2.13 \end{array} $$ a. Let \(\mu_{1}\) be the mean of population 1 and \(\mu_{2}\) be the mean of population \(2 .\) What is the point estimate of \(\mu_{1}-\mu_{2} ?\) b. Construct a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Test at the \(2.5 \%\) significance level if \(\mu_{1}\) is lower than \(\mu_{2}\).

A new type of sleeping pill is tested against an older, standard pill. Two thousand insomniacs are randomly divided into two equal groups. The first group is given the old pill, and the second group receives the new pill. The time required to fall asleep after the pill is administered is recorded for each person. The results of the experiment are given in the following table, where \(\bar{x}\) and \(s\) represent the mean and standard deviation, respectively, for the times required to fall asleep for people in each group after the pill is taken. $$ \begin{array}{lcc} \hline & \begin{array}{c} \text { Group 1 } \\ \text { (Old PilI) } \end{array} & \begin{array}{c} \text { Group 2 } \\ \text { (New Pill) } \end{array} \\ \hline n & 1000 & 1000 \\ \bar{x} & 15.4 \text { minutes } & 15.0 \text { minutes } \\ s & 3.5 \text { minutes } & 3.0 \text { minutes } \\ \hline \end{array} $$ Consider the test of hypothesis \(H_{0}: \mu_{1}-\mu_{2}=0\) versus \(H_{1}: \mu_{1}-\mu_{2}>0\), where \(\mu_{1}\) and \(\mu_{2}\) are the mean times required for all potential users to fall asleep using the old pill and the new pill, respectively. a. Find the \(p\) -value for this test. b. Does your answer to part a indicate that the result is statistically significant? Use \(\alpha=.025\). c. Find the \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). d. Does your answer to part c imply that this result is of great practical significance?
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