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Chapter 10: Problem 27

A Bloomberg News poll found that Americans plan to keep spending down over the next six months due to the uncertain economy (USA Today, September 17, 2009). Suppose a group of 15 households noted their household spending in March and then noted their household spending six months later in September. The mean monthly difference (former spending - current spending was calculated to be \(\$ 75.50\) with a standard deviation of \(\$ 66.20 .\) Does this sample of households show sufficient evidence of increased household savings? Use the 0.05 level of significance and assume normality of spending amounts.

Short Answer

Expert verified
No, there is not sufficient evidence at the 0.05 level of significance to conclude that household savings have increased.

Step by step solution

01

Stating the Hypotheses

The first step in hypothesis testing is to state the null hypothesis and the alternative hypothesis. The null hypothesis is that the mean difference in spending is equal to \$75.50, which corresponds to no change in savings. The alternative hypothesis is that the mean difference in spending is more than \$75.50, corresponding to an increase in savings.So, we can write the null hypothesis (\(H_0\)) as: \(\mu=75.50\) and the alternative hypothesis (\(H_a\)): \(\mu > 75.50\)
02

Compute the Test Statistic

Because we do not know the population standard deviation, we will use the t-test statistic. The formula for the t-statistic is: \(t = \frac{(\overline{X}- \mu_{0})}{(s/\sqrt{n})}\), where \(\overline{X}\) is the sample mean, \(\mu_{0}\) is the hypothesized population mean, \(s\) is the sample standard deviation and \(n\) is the sample size. Substituting the provided values into the formula we get: \(t = \frac{(75.50 - 75.50)}{(66.20/\sqrt{15})}\), which simplifies to \(t=0\).
03

Determine the Critical Value and Make a Decision

We look up the critical value for a one-tailed t-test with \(n-1=14\) degrees of freedom and \(\alpha=5\%=0.05\) level of significance. The critical value is approximately 1.761. Because our calculated t-statistic (0) is less than the critical value, we do not reject the null hypothesis.

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Key Concepts

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

t-test statistic
The **t-test statistic** is a crucial tool in hypothesis testing, especially when dealing with small sample sizes. It helps test the mean of a small sample against a known value when the population standard deviation is unknown. In the context of the exercise, we wanted to check if there's a significant difference in spending over time.To compute the t-test statistic, we use the formula:\[ t = \frac{(\overline{X} - \mu_{0})}{(s/\sqrt{n})} \]Here:
  • \( \overline{X} \) is the sample mean of the differences, which is 75.50 in our example.
  • \( \mu_{0} \) is the hypothesized population mean.
  • \( s \) is the sample standard deviation.
  • \( n \) is the number of samples.
The aim is to determine if our sample mean significantly deviates from \( \mu_{0} \). In our example, the calculated t-statistic was 0, which indicates no significant difference from the hypothesized mean.
null hypothesis
In hypothesis testing, the **null hypothesis** denoted as \( H_0 \), is the statement we are testing. It's the default or original assumption and often implies no effect or no difference.In our exercise, the null hypothesis was:\[ H_0: \mu = 75.50 \]This means that there is no significant increase in household savings, effectively stating what's ordinary or already expected. The mean difference in spending is exactly \$75.50.Testing the null hypothesis involves trying to find evidence in the data that it's not true. If our sample provides enough evidence to contradict this hypothesis, we might reject it in favor of the alternative.
alternative hypothesis
The **alternative hypothesis**, denoted as \( H_a \), is what you might conclude if you reject the null hypothesis. It represents the outcome you anticipate or suspect to be true.For the exercise in question, the alternative hypothesis was:\[ H_a: \mu > 75.50 \]This suggests that there is an increase in household savings, meaning the mean spending difference is greater than \$75.50.Unlike the null hypothesis, which reflects skepticism about a significant change, the alternative hypothesis is optimistic or hopeful about a meaningful change or effect. If evidence is strong enough, rejecting \( H_0 \) allows us to accept this alternative proposition confidently.
critical value
The **critical value** is used to decide whether to reject the null hypothesis. In our context, it represents the boundary above or below which we consider evidence significant enough to reject \( H_0 \).To find the critical value:
  • Determine the level of significance, \( \alpha \), which was 0.05 in our example.
  • Find the degrees of freedom, which is \( n-1 \) for the t-test, counted as 14 for our sample size of 15.
  • Use a t-distribution table or calculator to find the critical value for a one-tailed test.
In the exercise, the critical value was approximately 1.761.If our t-test statistic equals or exceeds this critical value, we reject \( H_0 \). Since the calculated t-value was 0, less than 1.761, we did not reject the null hypothesis, signifying no strong evidence of increased household savings.

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Most popular questions from this chapter

State the null hypothesis, \(H_{o},\) and the alternative hypothesis, \(H_{a},\) that would be used to test these claims: a. The mean of the differences between the post-test and the pre-test scores is greater than \(15 .\) b. The mean weight gain, after the change in diet for the laboratory animals, is at least 10 oz. c. The mean weight loss experienced by people on a new diet plan was no less than 12 lb. d. The mean difference in the home reassessments from the two town assessors was no more than \(\$ 200\)

Determine the \(p\) -value that would be used to test the following hypotheses when \(F\) is used as the test statistic: a. \( H_{o}: \sigma_{1}=\sigma_{2}\) versus \(H_{a}: \sigma_{1}>\sigma_{2},\) with \(n_{1}=10, n_{2}=16\) and \(F_{*}=2.47\) b. \(H_{o}: \sigma_{1}^{2}=\sigma_{2}^{2}\) versus \(H_{a}: \sigma_{1}^{2}>\sigma_{2}^{2},\) with \(n_{1}=25, n_{2}=21\) and \(F *=2.31\) c. \(H_{o}: \frac{\sigma_{1}^{2}}{\sigma_{2}^{2}}=1\) versus \(H_{a}: \frac{\sigma_{1}^{2}}{\sigma_{2}^{2}} \neq 1,\) with \(n_{1}=41, n_{2}=61\) and \(F \star=4.78\) d. \( H_{o}: \sigma_{1}=\sigma_{2}\) versus \(H_{a}: \sigma_{1}<\sigma_{2},\) with \(n_{1}=10, n_{2}=16\) and \(F_{*}=2.47\)

State the null hypothesis, \(H_{o}\), and the alternative hypothesis, \(H_{a},\) that would be used to test the following claims: a. The standard deviation of population \(X\) is smaller than the standard deviation of population Y. b. The ratio of the variances of population A over population \(B\) is greater than 1. c. The standard deviation of population \(Q_{1}\) is at most that of population \(\mathrm{Q}_{2}\) d. The variability within population I is more than the variability within population II.

Find the confidence coefficient, \(t\left(\mathrm{df}, \frac{\alpha}{2}\right),\) that would be used to find the maximum error for each of the following situations when estimating the difference between two means, \(\mu_{1}-\mu_{2}\) a. \(\quad 1-\alpha=0.95, n_{1}=25, n_{2}=15\) b. \(1-\alpha=0.98, n_{1}=43, n_{2}=32\) c. \(\quad 1-\alpha=0.99, n_{1}=19, n_{2}=45\)

A study in the New England Journal of Medicine reported that based on 987 deaths in southern California, right-handers died at an average age of 75 and left-handers died at an average age of \(66 .\) In addition, it was found that \(7.9 \%\) of the lefties died from accident-related injuries, excluding vehicles, versus \(1.5 \%\) for the right-handers; and \(5.3 \%\) of the left- handers died while driving vehicles versus \(1.4 \%\) of the right-handers. Suppose you examine 1000 randomly selected death certificates, of which 100 were left-handers and 900 were right-handers. If you found that 5 of the left- handers and 18 of the right-handers died while driving a vehicle, would you have evidence to show that the proportion of left-handers who die at the wheel is significantly higher than the proportion of right-handers who die while driving? Calculate the \(p\) -value and interpret its meaning.
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