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The mean balance of all checking accounts at a bank on December 31,2009, 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 the \(1 \%\) 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
Depends on whether the calculated t statistic is less than the critical t value at the given significance level. If it is, there is enough evidence to conclude the mean balance has decreased. If it's not, there is not enough evidence to suggest a decrease in the mean balance.

Step by step solution

01

Identify the Null and Alternate Hypothesis

Null Hypothesis, \(H_0\): The mean balance of bank accounts has NOT decreased. \nAlternative Hypothesis, \(H_a\): The mean balance of bank accounts has decreased.
02

Calculate the t statistic

The t statistic is the standardized difference between the sample mean and the population mean. It can be calculated with this formula: \( t = \(\frac{\bar{x} - \mu}{s/ \sqrt{n}} \) \nWhere, \(\bar{x}\) = sample mean = $780, \(\mu\) = population mean = $850, \(s\) = standard deviation = $230 and \(n\) the number of observations = 55. After substituting the figures, calculate t.
03

Determine the critical t value at \(1\%\) level

Look up the critical t value for a one-tailed test at the \(1\%\) significance level with \(n - 1= 55 - 1 = 54\) degrees of freedom from a t distribution table. Compare the calculated t statistic to this critical t value.
04

Determine the critical t value at \(2.5\%\) level

Look up the critical t value for a one-tailed test at the \(2.5\%\) significance level with \(54\) degrees of freedom from a t distribution table. Compare the calculated t statistic to this critical t value.
05

Draw conclusion based on t statistic and critical t values

If the calculated t statistic is less than the critical t value at either significance level, reject the null hypothesis. Conclude that the there is sufficient evidence, at that specified level of significance, to suggest the mean balance of bank accounts decreased. Conversely, if the calculated t statistic is not less than the critical t value, don't reject the null hypothesis. Conclude that there is not sufficient evidence to suggest the mean balance has decreased.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis is an important concept. It represents a default position that there is no effect or change. In our example, the null hypothesis (\( H_0 \)) states that the mean balance of the checking accounts has not decreased from the previous mean balance of $850. The null hypothesis basically acts like a baseline assumption. We assume \( H_0 \) is true until we have enough evidence to suggest otherwise.Understanding \( H_0 \) is crucial because it helps define what we are testing. It helps us understand what measurements or observations would "prove" our alternative claim. Essentially, we test against this baseline to see if our data shows a different story.A null hypothesis often involves statements like "no effect," "no difference," or, as in this case, "no decrease." It's the opposite of what researchers are usually trying to prove.
Alternative Hypothesis
The alternative hypothesis is the statement we are trying to support with our statistical analysis. In the context of the problem, the alternative hypothesis (\( H_a \)) suggests that the mean balance of the checking accounts has decreased.This hypothesis stands "in opposition" to the null hypothesis. We aren't just looking for any decrease but a statistically significant decrease. This means any observed decrease must exceed random chance occurrence; it should be provable by numbers.Alternative hypotheses often reflect what researchers aim to demonstrate - that there is a change, effect, or difference to consider. In terms of proving it, the process involves demonstrating through data analysis (e.g., calculating t statistics) that rejecting the null hypothesis in favor of the alternative is justified.In simpler terms, if evidence supports \( H_a \), then the decline is statistically significant.
T Statistic
The t statistic is a value calculated from our sample data during a t-test to measure the size of the difference relative to the variation in our sample data. We use the formula:\[t = \frac{\bar{x} - \mu}{s/ \sqrt{n}}\]Where:- \( \bar{x} \) is the sample mean,- \( \mu \) is the population mean,- \( s \) is the sample standard deviation,- \( n \) is the number of observations.In our example, the sample mean is \(780, the population mean is \)850, with a standard deviation of $230 over 55 observations.Calculating the t statistic helps us compare our sample results against the null hypothesis. If this t statistic shows a large enough "difference," it could suggest that the true mean balance has indeed decreased. The larger the t statistic, the greater evidence against the null hypothesis.
Significance Level
The significance level, denoted as \( \alpha \), is the criterion used for rejecting the null hypothesis. It represents the probability of rejecting a true null hypothesis.In simpler terms, it's the chance of making a mistake if we think there is a decrease in the mean balance, but there actually isn't.- At a 1% significance level, or \( \alpha = 0.01 \), we're only willing to incorrectly reject \( H_0 \) 1% of the time. This indicates high confidence.- At a 2.5% significance level, or \( \alpha = 0.025 \), the confidence is slightly lower, as we're willing to make such a mistake 2.5% of the time.Choosing a lower \( \alpha \) means we'll need stronger evidence to reject \( H_0 \). It defines the "cutoff point" for deciding if \( H_0 \) should be rejected in favor of \( H_a \).
T Distribution
The t distribution is a type of probability distribution that is symmetric and bell-shaped, similar to the normal distribution but with heavier tails. It is used when estimating population parameters when the sample size is small, and the population standard deviation is unknown.The shape of the t distribution is determined by the degrees of freedom, which in our case is calculated as \( n-1 \), where \( n \) is the sample size. As the sample size increases, the t distribution approaches the normal distribution.In hypothesis testing, the t distribution helps us determine the critical t values - the threshold values - to decide whether to reject \( H_0 \). We compare our calculated t statistic to the critical t value from the t distribution table, based on our chosen significance level and degrees of freedom.The t distribution is crucial for determining the "boundaries" of expected variability in our test statistic.

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Most popular questions from this chapter

At Farmer's Dairy, a machine is set to fill 32 -ounce milk cartons. However, this machine does not put exactly 32 ounces of milk into each carton; the amount varies slightly from carton to carton but has a normal distribution. It is known that when the machine is working properly, the mean net weight of these cartons is 32 ounces. The standard deviation of the milk in all such cartons is always equal to \(.15\) ounce. The quality control inspector at this company takes a sample of 25 such cartons every week, calculates the mean net weight of these cartons, and tests the null hypothesis, \(\mu=32\) ounces, against the alternative hypothesis, \(\mu \neq 32\) ounces. If the null hypothesis is rejected, the machine is stopped and adjusted. A recent sample of 25 such cartons produced a mean net weight of \(31.93\) ounces a. Calculate the \(p\) -value for this test of hypothesis. Based on this \(p\) -value, will the quality control inspector decide to stop the machine and readjust it if she chooses the maximum probability of a Type I error to be \(.01 ?\) What if the maximum probability of a Type I error is .05? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.01\). Does the machine need to be adjusted? What if \(\alpha=.05 ?\)

Consider the null hypothesis \(H_{0}: p=.65\). Suppose a random sample of 1000 observations is taken to perform this test about the population proportion. Using \(\alpha=.05\), show the rejection and nonrejection regions and find the critical value(s) of \(z\) for a a. left-tailed test b. two-tailed test c. right-tailed test

Perform the following tests of hypothesis. a. \(H_{0}: \mu=285, \quad H_{1}: \mu<285, \quad n=55, \bar{x}=267.80, \quad s=42.90, \quad \alpha=.05\) b. \(H_{0}: \mu=10.70, \quad H_{1}: \mu \neq 10.70, \quad n=47, \bar{x}=12.025, \quad s=4.90, \quad \alpha=.01\) c. \(H_{0}: \mu=147,500, \quad H_{1}: \mu<147,500, n=41, \bar{x}=141,812, s=22,972, \alpha=.10\)

In Las Vegas and Atlantic City, New Jersey, tests are performed often on the various gaming devices used in casinos. For example, dice are often tested to determine if they are balanced. Suppose you are assigned the task of testing a die, using a two-tailed test to make sure that the probability of a 2 -spot is \(1 / 6 .\) Using the \(5 \%\) significance level, determine how many 2 -spots you would have to obtain to reject the null hypothesis when your sample size is \(\begin{array}{lll}\text { a. } 120 & \text { b. } 1200 & \text { c. } 12,000\end{array}\) Calculate the value of \(\hat{p}\) for each of these three cases. What can you say about the relationship between (1) the difference between \(\hat{p}\) and \(1 / 6\) that is necessary to reject the null hypothesis and (2) the sample size as it gets larger?

A 2008 AARP survey reported that \(85 \%\) of U.S. workers aged 50 years and older with at least one 4-year college degree had taken employer-based training within the previous 2 years, compared to only \(50 \%\) of workers aged 50 years and older with a high school degree or less. In a current survey of \(640 \mathrm{U.S}\). workers aged 50 years and older with a high school degree or less, 341 had taken employer-based training within the previous 2 years. a. Using the critical-value approach and \(\alpha=.05\), test whether the current percentage of all U.S. workers aged 50 years and older with a high school degree or less who have taken employerbased training within the previous 2 years is different from \(50 \%\). b. How do you explain the Type I error in part a? What is the probability of making this error in part a? c. Calculate the \(p\) -value for the test of part a. What is your conclusion if \(\alpha=.05 ?\)

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