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The melting point of each of 16 samples of a brand of hydrogenated vegetable oil was determined, resulting in \(\bar{x}=94.32\). Assume that the distribution of melting point is normal with \(\sigma=1.20\). a. Test \(H_{0}: \mu=95\) versus \(H_{\mathrm{a}}: \mu \neq 95\) using a two- tailed level .01 test. b. If a level \(.01\) test is used, what is \(\beta(94)\), the probability of a type II error when \(\mu=94\) ? c. What value of \(n\) is necessary to ensure that \(\beta(94)=.1\) when \(\alpha=.01 ?\)

Short Answer

Expert verified
a. Do not reject \(H_0\). b. \(\beta(94) \approx 0.5733\). c. Sample size \(n = 34\).

Step by step solution

01

State the null and alternative hypothesis

In part (a), we are testing the null hypothesis \( H_0: \mu = 95 \) against the alternative hypothesis \( H_a: \mu eq 95 \). This is a two-tailed test at a significance level of \( \alpha = 0.01 \).
02

Calculate the test statistic

To calculate the test statistic for a z-test, use the formula:\[Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}\]where \( \bar{x} = 94.32 \), \( \mu = 95 \), \( \sigma = 1.20 \), and \( n = 16 \). Substitute the values in:\[Z = \frac{94.32 - 95}{1.20 / \sqrt{16}} = \frac{-0.68}{0.30} \approx -2.27\]
03

Find critical values and decision rule

For a two-tailed test with \( \alpha = 0.01 \), the critical z-values are \( \pm 2.576 \). If the test statistic falls outside this range, we reject the null hypothesis.
04

Make a decision based on the test statistic

The test statistic calculated is \( -2.27 \), which does not fall outside the range \(-2.576, 2.576\). Hence, we do not reject the null hypothesis \( H_0 \).
05

Calculate the probability of a Type II error for \( \mu = 94 \)

To find \( \beta \) when \( \mu = 94 \), calculate the z-values corresponding to the critical values:\[z_1 = \frac{94.32 - 94}{1.20 / \sqrt{16}} = \frac{0.32}{0.30} = 1.067\]\[z_2 = \frac{93.68 - 94}{1.20 / \sqrt{16}} = \frac{-0.32}{0.30} = -1.067\]Find the probability that the calculated z-values are between these critical z-values, using the normal distribution table. This gives:\[\beta(94) = P(-1.067 < Z < 1.067) = 2 \times P(Z < 1.067) - 1 \approx 0.5733\]
06

Determine necessary sample size n for \( \beta(94) = 0.1 \)

With \( \alpha = 0.01 \) and \( \beta = 0.1 \), use the formula:\[n = \left( \frac{(Z_{\alpha/2} + Z_{\beta}) \cdot \sigma}{\Delta} \right)^2\]where \( Z_{\alpha/2} = 2.576 \), \( Z_{\beta} \approx 1.28 \), \( \sigma = 1.20 \), and \( \Delta = 1.0 \). Thus,\[n = \left( \frac{(2.576 + 1.28) \cdot 1.20}{1.0} \right)^2 = \left( \frac{4.856 \cdot 1.20}{1.0} \right)^2 = \left( 5.828 \right)^2 \approx 33.97 \approx 34\]Therefore, a sample size of 34 is needed.

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Key Concepts

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

Normal Distribution
When we talk about the normal distribution, we are referring to a continuous probability distribution that is symmetrical around its mean. This distribution is often referred to as a bell curve due to its distinctive shape. In statistics, the normal distribution is crucial because many variables are naturally distributed in this way.

Some key features include:
  • Mean, median, and mode of the distribution are all equal.
  • The curve is symmetric around the mean.
  • The total area under the curve is 1.
  • About 68% of the data falls within one standard deviation of the mean, 95% within two, and 99.7% within three.
In the context of hypothesis testing, particularly when conducting a z-test, it is often assumed that the data follows a normal distribution. This assumption helps statisticians make inferences about the population parameters using sample statistics.
Z-Test
The z-test is a statistical test used to determine whether there is a significant difference between the mean of a sample and the mean of a population from which the sample is drawn. It is particularly useful when the population variance is known and the sample size is sufficiently large (typically \(n > 30\)).

In our scenario, even though the sample size is 16, we assume that the normality condition is met due to the problem's assumptions. Here are the steps generally taken:
  • Calculate the z-score using the formula: \[Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}\]
    • \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the standard deviation, and \(n\) is the sample size.
  • Determine the critical value for the desired significance level \(\alpha\), which marks the threshold for rejecting the null hypothesis.
  • Evaluate whether the calculated z-score falls in the critical region to decide if the null hypothesis should be rejected.
If the z-score falls beyond the critical values, the null hypothesis is rejected, suggesting that the sample provides enough evidence that the population mean is different from the hypothesized mean.
Type II Error
A Type II error occurs in hypothesis testing when the null hypothesis is not rejected when it is actually false. It is essentially a false negative and is denoted by \(\beta\).

This type of error can have implications, particularly where failing to identify a true effect can be costly.
  • Factors that can impact \(\beta\) include:
    • Sample size: Larger samples tend to have smaller \(\beta\) values.
    • Significance level \(\alpha\): A lower significance level increases \(\beta\).
    • Effect size: Larger true differences between the sample and population mean reduce \(\beta\).
To calculate \(\beta\) for a specific mean, as shown in our problem, comparison is made between the theoretical distribution and standard normal distribution, using calculated z-values. The calculated probability then informs the risk of making a Type II error under specific conditions. In such calculations, the software or z-table is often utilized for accuracy and efficiency.
Sample Size Determination
Sample size determination is a critical aspect of experimental design. It involves calculating the number of observations or samples required to achieve a specified level of accuracy.

For hypothesis testing, determining the right sample size ensures the test's effectiveness by balancing between Type I and Type II errors. The formula to find the necessary sample size, given a specific \(\beta\) and \(\alpha\), is:\[n = \left( \frac{(Z_{\alpha/2} + Z_{\beta}) \cdot \sigma}{\Delta} \right)^2\]where:
  • \(\Delta\) is the smallest effect size of interest.
  • \(Z_{\alpha/2}\) is the critical value for the significance level.
  • \(Z_{\beta}\) is the z-score corresponding to the desired \(\beta\).
  • \(\sigma\) is the population standard deviation.
In our exercise, calculating \(n\) ensures that a specific power of the test \((1 - \beta)\) is achieved, which indicates a high probability of correctly rejecting the null hypothesis when it is indeed false. This balance between errors is crucial to making reliable inferences in hypothesis testing.

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Most popular questions from this chapter

The article "Caffeine Knowledge, Attitudes, and Consumption in Adult Women" \((J\). Nutrit. Ed., 1992: 179-184) reports the following summary data on daily caffeine consumption for a sample of adult women: \(n=47, \bar{x}=215 \mathrm{mg}, s=235\) \(\mathrm{mg}\), and range \(=5-1176\). a. Does it appear plausible that the population distribution of daily caffeine consumption is normal? Is it necessary to assume a normal population distribution to test hypotheses about the value of the population mean consumption? Explain your reasoning. b. Suppose it had previously been believed that mean consumption was at most \(200 \mathrm{mg}\). Does the given data contradict this prior belief? Test the appropriate hypotheses at significance level \(.10\) and include a \(P\)-value in your analysis.

Pairs of \(P\)-values and significance levels, \(\alpha\), are given. For each pair, state whether the observed \(P\) value would lead to rejection of \(H_{0}\) at the given significance level. a. \(P\)-value \(=.084, \alpha=.05\) b. \(P\)-value \(=.003, \alpha=.001\) c. \(P\)-value \(=.498, \alpha=.05\) d. \(P\)-value \(=.084, \alpha=.10\) e. \(P\)-value \(=.039, \alpha=.01\) f. \(P\)-value \(=.218, \alpha=.10\)

A hot-tub manufacturer advertises that with its heating equipment, a temperature of \(100^{\circ} \mathrm{F}\) can be achieved in at most \(15 \mathrm{~min}\). A random sample of 32 tubs is selected, and the time necessary to achieve a \(100^{\circ} \mathrm{F}\) temperature is determined for each tub. The sample average time and sample standard deviation are \(17.5 \mathrm{~min}\) and \(2.2 \mathrm{~min}\), respectively. Does this data cast doubt on the company's claim? Compute the \(P\)-value and use it to reach a conclusion at level \(.05\) (assume that the heating-time distribution is approximately normal).

A pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least \(10 \mathrm{~h}\). A random sample of 18 pens is selected, the writing lifetime of each is determined, and a normal probability plot of the resulting data supports the use of a one-sample \(t\) test. a. What hypotheses should be tested if the investigators believe a priori that the design specification has been satisfied? b. What conclusion is appropriate if the hypotheses of part (a) are tested, \(t=-2.3\), and \(\alpha=.05 ?\) c. What conclusion is appropriate if the hypotheses of part (a) are tested, \(t=-1.8\), and \(\alpha=.01 ?\) d. What should be concluded if the hypotheses of part (a) are tested and \(t=-3.6\) ?

Because of variability in the manufacturing process, the actual yielding point of a sample of mild steel subjected to increasing stress will usually differ from the theoretical yielding point. Let \(p\) denote the true proportion of samples that yield before their theoretical yielding point. If on the basis of a sample it can be concluded that more than \(20 \%\) of all specimens yield before the theoretical point, the production process will have to be modified. a. If 15 of 60 specimens yield before the theoretical point, what is the \(P\)-value when the appropriate test is used, and what would you advise the company to do? b. If the true percentage of "early yields" is actually \(50 \%\) (so that the theoretical point is the median of the yield distribution) and a level \(.01\) test is used, what is the probability that the company concludes a modification of the process is necessary?

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