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Chapter 9: Problem 61

The mean balance of all checking accounts at a bank on December 31,2011, was \(\$ 850 .\) A random sample of 55 checking accounts taken recently from this bank gave a mean balance of \(\$ 780\) with a standard deviation of \(\$ 230 .\) Using a \(1 \%\) significance level, can you conclude that the mean balance of such accounts has decreased during this period? Explain your conclusion in words. What if \(\alpha=.025\) ?

Short Answer

Expert verified
To determine whether the mean balance of accounts has decreased, a hypothesis test is conducted. If the calculated Z score is less than the negative critical value, the null hypothesis is rejected, and it can be concluded that the mean balance has decreased. These comparisons are done twice, once for a 1% significance level and again for a 2.5% significance level.

Step by step solution

01

State the Hypotheses

The null hypothesis (\(H_0\)): The mean balance is $850, i.e., \( \mu = 850\). The alternative hypothesis (\(H_1\)): The mean balance has decreased, i.e., \( \mu < 850\)
02

Calculate the Test Statistic

Calculate the Z score using the formula \(Z = \frac{{\bar{x} - \mu}}{{\sigma / \sqrt{n}}}\), where \(\bar{x}\) is sample mean, \(\mu\) is population mean, \(\sigma\) is standard deviation and \(n\) is the number of samples. So, \(Z = \frac{{780 - 850}}{{230 / \sqrt{55}}}\)
03

Find the Critical Value

The negative critical value (Z critical) for a 1% significance level in a one-tailed test can be obtained from a standard normal distribution table. For a 1% significance level, the critical value is -2.33.
04

Make the Decision

If the Z score calculated in step 2 is less than -2.33, then we reject the null hypothesis. Otherwise, we do not reject it.
05

What if \(\alpha=.025\)

If \(\alpha=.025\), then it represents a 2.5% significance level. We still conduct a one-tail test, but with a new critical value corresponding to the new significance level. Using a standard normal distribution table, this new critical value is approximately -1.96. Repeating step 4 with this new critical value allows to make the decision about the null hypothesis based on this new significance level.
06

Explaining the Conclusion

The decision made in step 4 and 5 can now be communicated. If the null hypothesis was rejected, it means there is enough statistical evidence to say that the mean balance has decreased at the given significance level. If the null hypothesis was not rejected, it simply means there isn't enough statistical evidence to say that the mean has decreased at the given significance level. It does not imply that the mean balance hasn't decreased.

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Key Concepts

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

Null Hypothesis
The null hypothesis (\( H_0 \)) is a fundamental part of hypothesis testing in statistics. It is a statement that assumes there is no effect or no difference, and in this context, it means the mean balance of the checking accounts has not decreased.
  • Simply put, it suggests "status quo" or no change from what we initially believe.
  • For our problem, the null hypothesis is that the mean balance is still $850, or \( \mu = 850 \) .
We use the null hypothesis as a starting point to test whether there is enough evidence to support a change or difference.
Alternative Hypothesis
In contrast, the alternative hypothesis (\( H_1 \)) proposes that there is an effect or a difference. This is what you aim to support with your data.
  • For the bank's checking account problem, the alternative hypothesis states that the mean balance has indeed decreased, i.e., \( \mu < 850 \)
It represents the new claim or the new idea we want to test and thus, challenges the null hypothesis.
Significance Level
The significance level, often denoted by \( \alpha \), is the threshold at which we decide whether to reject the null hypothesis. It's essentially the probability of rejecting the null hypothesis when it is true.
  • Common significance levels are 0.05, 0.01, or 0.025, depending on how conclusive you want your results.
  • In our testing scenario, a 1\(\%\) significance level (\( \alpha = 0.01 \)) was initially chosen.
The lower the significance level, the stronger the evidence must be to reject the null hypothesis, making it harder to declare a statistical difference.
Z-Score
The Z-score is a statistical measure that helps us understand how far away a sample statistic is from the null hypothesis population parameter. In our example, it helps compare the sample mean with the expected mean under the null hypothesis.
  • The formula for the Z-score is \( Z = \frac{{\bar{x} - \mu}}{{\sigma / \sqrt{n}}} \), where \( \bar{x} \) is the sample mean and \( \mu \) is the population mean.
  • It indicates how many standard deviations the sample mean is away from the null hypothesis mean.
The Z-score helps us make the decision about rejecting or not rejecting the null hypothesis.
Critical Value
A critical value is a point on the test distribution that is compared to the test statistic to decide whether to reject the null hypothesis. It defines the boundary for the critical region, beyond which we consider the results statistically significant.
  • For a 1\(\%\) significance level in our example, the critical value is \(-2.33\).
  • The boundaries change if you use a different significance level, like \( \alpha = 0.025 \), where the critical value would be approximately \(-1.96\).
The critical value acts as a cutoff point: if our calculated statistic is more extreme than the critical value, we reject the null hypothesis.

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Most popular questions from this chapter

Records in a three-county area show that in the last few years. Girl Scouts sold an average of \(47.93\) boxes of cookies per year per girl scout, with a population standard deviation of \(8.45\) boxes per year. Fifty randomly selected Girl Scouts from the region sold an average of \(46.54\) boxes this year. Scout leaders are concerned that the demand for Girl Scout cookies may have decreased A. Test at a \(10 \%\) significance level whether the average number of boxes of cookies sold by all Gir Scouts in the three-county area is lower than the historical average of \(47.92\) b. What will your decision be in part a if the probability of a Type I error is zero? Expl.

A business school claims that students who complete a 3-month typing course can type, on average at least 1200 words an hour. A random sample of 25 students who completed this course typed, on aver 1125 words an hour with a standard deviation of 85 words. Assume that the typing speeds for all stu lents. who complete this course have an approximate normal distribution Suppose the probability of making a Type I error is selected to be zero. Can you conclude that the claim of the business school is true? Answer without performing the five steps of a test of hypothe b. Using a \(5 \%\) significance level, can you conclude that the claim ousiness school is true? \(\mathrm{Us}\) both approaches

Briefly explain the meaning of each of the following terms. a. Null hypothesis b. Alternative hypothesis c. Critical point(s) d. Significance level e. Nonrejection region f. Rejection region g. Tails of a test h. Two types of errors

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According to a New York Times/CBS News poll conducted during June \(24-28,2011,55 \%\) of the American adults polled said that owning a home is a very important part of the American Dream (The New York Times, June 30,2011 ). Suppose this result was true for the population of all American adults in \(2011 .\) In a recent poll of 1800 American adults, \(61 \%\) said that owning a home is a very important part of the American Dream. Perform a hypothesis test to determine whether it is reasonable to conclude that the percentage of all American adults who currently hold this opinion is higher than \(55 \%\). Use a \(2 \%\) significance level, and use both the \(p\) -value and the critical-value approaches.
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