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Perform the indicated test of hypotheses, based on the information given. a. Test \(H_{0: \mu}=105\) vs. Ha: \(\mu>105 @ \alpha=0.05, \sigma\) unknown, \(n=30, x-=108, s=7.2\) b. Test \(H 0: \mu=21.6\) vs. Ha: \(\mu<21.6 @ \alpha=0.01, \sigma\) unknown, \(n=78, x-=20.5, s=3.9\) c. Test \(H 0: \mu=-375\) VS. Ha: \(\mu \neq-375 @ \alpha=0.01, \sigma=18.5, n=31, x-=-388, s=18.0\)

Short Answer

Expert verified
(a) Reject \(H_0\). (b) Do not reject \(H_0\). (c) Reject \(H_0\).

Step by step solution

01

Establish Hypotheses and Identify Parameters

For problem (a), identify the null hypothesis, \(H_0: \mu = 105\), and alternative hypothesis, \(H_a: \mu > 105\). For problem (b), \(H_0: \mu = 21.6\), and \(H_a: \mu < 21.6\). For problem (c), \(H_0: \mu = -375\), and \(H_a: \mu eq -375\). Note the sample sizes, means, and standard deviations, and determine the significance levels: 0.05 for (a), 0.01 for (b), and 0.01 for (c).
02

Select the Appropriate Test and Calculate Test Statistic

For problems (a) and (b), use a t-test since \(\sigma\) is unknown. For problem (c), with \(\sigma\) known, use a z-test.(a) \(t = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{108 - 105}{7.2/\sqrt{30}} = 2.282\).(b) \(t = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{20.5 - 21.6}{3.9/\sqrt{78}} = -2.090\).(c) \(z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} = \frac{-388 + 375}{18.5/\sqrt{31}} = -3.736\).
03

Determine Critical Value and Decision Rule

Use the t-distribution table for problems (a) and (b) and the z-table for problem (c).(a) For a one-tailed test at \(\alpha = 0.05\) and \(df = 29\), the critical t-value is approximately 1.699.(b) For a one-tailed test at \(\alpha = 0.01\) and \(df = 77\), the critical t-value is approximately -2.639.(c) For a two-tailed test at \(\alpha = 0.01\), the critical z-value is approximately \(\pm 2.576\).
04

Compare Test Statistic to Critical Value

Compare computed test statistics to critical values for decision making.(a) The test statistic, 2.282, is greater than the critical value, 1.699; thus, reject \(H_0\).(b) The test statistic, -2.090, is not less than the critical value, -2.639; do not reject \(H_0\).(c) The test statistic, -3.736, is less than the critical value, -2.576; therefore, reject \(H_0\).
05

State Conclusion

Summarize the conclusion based on the comparison. (a) There is sufficient evidence at the 0.05 significance level to conclude that the mean is greater than 105. (b) There is not enough evidence at the 0.01 significance level to conclude that the mean is less than 21.6. (c) There is sufficient evidence at the 0.01 significance level to conclude that the mean is not equal to -375.

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Key Concepts

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

t-test
The t-test is a fundamental statistical tool often used when the population standard deviation, \( \sigma \), is unknown, and the sample size is small. This test helps determine if there is a significant difference between the mean of a sample and a known or hypothesized population mean. In hypothesis testing, a t-test is chosen if the sample distribution is approximately normally distributed, which becomes especially important with smaller sample sizes (generally if \( n < 30 \)).
One critical element of the t-test is its reliance on the sample standard deviation, \( s \), instead of the population standard deviation, which can be unknown. The formula for the t-statistic is:
  • \[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]
Here, \( \bar{x} \) is the sample mean, \( \mu \) is the population mean hypothesis, \( s \) is the sample standard deviation, and \( n \) is the sample size. The degrees of freedom used in the t-test is \( n - 1 \).
In hypothesis testing problems, such as exercises (a) and (b) where \( \sigma \) was unknown, the t-test was the appropriate choice due to its adaptability to sample variances and smaller sample sets.
z-test
The z-test is another essential method in hypothesis testing, typically used when the population standard deviation is known and the sample size is large (usually \( n \geq 30 \)). It is applicable for comparing a sample mean to a population mean when the data is normally distributed.
In a z-test, the calculated z-score helps us understand how far away our sample mean is from the population mean, in terms of the population standard deviation. The formula for calculating the z-statistic is:
  • \[ z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \]
Where \( \bar{x} \) is the sample mean, \( \mu \) is the hypothesized population mean, \( \sigma \) is the population standard deviation, and \( n \) is the sample size. The assumption here is that the population data is normally distributed, which justifies using the z-test.
In exercise problem (c), the z-test was deemed suitable because the population standard deviation \( \sigma \) was known, thus allowing for the direct application of the z-test formula.
significance level
The significance level, often denoted by \( \alpha \), is a crucial concept in hypothesis testing. It represents the probability of rejecting a true null hypothesis, essentially indicating the risk we are willing to take of making a Type I error (false positive).
Common significance levels include 0.05, 0.01, and 0.10, and these thresholds are used to determine the critical values for our test statistics. It helps decide whether the findings are meaningful enough to reject the null hypothesis. A smaller \( \alpha \) indicates stronger evidence is needed to reject the null hypothesis.
For example, in exercise problems (a), (b), and (c), significance levels were set at 0.05 and 0.01. These decided how extreme the sample statistic needed to be in order to consider the results statistically significant. More conservatively, a 0.01 level, as used in problems (b) and (c), demands more substantial evidence to reject the hypothesis than a 0.05 level used in problem (a).
Choosing the right significance level depends on the field of study and the potential consequences of making a Type I error, with more critical fields often opting for lower significance level thresholds.

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Most popular questions from this chapter

The labor charge for repairs at an automobile service center are based on a standard time specified for each type of repair. The time specified for replacement of universal joint in a drive shaft is one hour. The manager reviews a sample of 30 such repairs. The average of the actual repair times is 0.86 hour with standard deviation 0.32 hour. a. Test at the \(1 \%\) level of significance the null hypothesis that the actual mean time for this repair differs from one hour. b. The sample mean is less than one hour, suggesting that the mean actual time for this repair is less than one hour. Perform this test, also at the \(1 \%\) level of significance. (The computation of the test statistic done in part (a) still applies here.)

Find the rejection region (for the standardized test statistic) for each hypothesis test. Identify the test as left-tailed, right-tailed, or two- tailed. a. \(\quad H 0: \mu=-62\) Vs. Ha: \(\mu \neq-62 @ \alpha=0.005\). b. \(\quad H_{0}: \mu=73\) VS. Ha: \(\mu>73 @ \alpha=0.001\). c. \(\quad H 0: \mu=1124\) VS. Ha: \(\mu<1124 @ \alpha=0.001\). d. \(\quad H 0: \mu=0.12\) VS. Ha: \(\mu \neq 0.12 @ \alpha=0.001\).

Authors of a computer algebra system wish to compare the speed of a new computational algorithm to the currently implemented algorithm. They apply the new algorithm to 50 standard problems; it averages 8.16 seconds with standard deviation 0.17 second. The current algorithm averages 8.21 seconds on such problems. Test, at the \(1 \%\) level of significance, the alternative hypothesis that the new algorithm has a lower average time than the current algorithm.

In the previous year the proportion of deposits in checking accounts at a certain bank that were made electronically was \(45 \%\). The bank wishes to determine if the proportion is higher this year. It examined 20,000 deposit records and found that 9,217 were electronic. Determine, at the \(1 \%\) level of significance, whether the data provide sufficient evidence to conclude that more than \(45 \%\) of all deposits to checking accounts are now being made electronically.

Compute the value of the test statistic for the indicated test, based on the information given. a. Testing \(H_{0}: \mu=342\) vs. \(H a: \mu<342, \sigma=11.2, n=40, x-=339, s=10.3\) b. Testing \(H_{0}: \mu=105\) vs. Ha: \(\mu>105, \sigma=5.3, n=80, x-=107, s=5.1\) c. Testing \(H_{0}: \mu=-13.5\) vs. Ha: \(\mu \neq-13.5, \sigma\) unknown, \(n=32, x-=-13.8, s=1.5\) d. Testing \(H_{0}: \mu=28\) vs. \(H a: \mu \neq 28, \sigma\) unknown, \(n=68, x-27.8, s=1.3\)

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