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Chapter 9: Problem 27

On the label, Pepperidge Farm bagels are said to weigh four ounces each ( \(113 \mathrm{~g}\) ). A random sample of six bagels resulted in the following weights (in grams): \(\begin{array}{llllll}117.6 & 109.5 & 111.6 & 109.2 & 119.1 & 110.8\end{array}\) a. Based on this sample, is there any reason to doubt that the population mean is at least \(113 \mathrm{~g}\) ? b. Assume that the population mean is actually \(110 \mathrm{~g}\) and that the distribution is normal with standard deviation \(4 \mathrm{~g}\). In a \(z\) test of \(H_{0}\) : \(\mu=113\) against \(H_{\mathrm{a}}: \mu<113\) with \(\alpha=.05\), find the probability of rejecting \(H_{0}\) with \(\operatorname{six}\) observations. c. Under the conditions of part (b) with \(\alpha=.05\), how many more observations would be needed in order for the power to be at least \(.95 ?\)

Short Answer

Expert verified
a. No, there is no reason. b. Probability is about 0.11. c. 20 observations needed.

Step by step solution

01

Calculate the Sample Mean

First, we calculate the sample mean (\( \bar{x} \)) of the given weights (117.6, 109.5, 111.6, 109.2, 119.1, 110.8 grams). Add all the sample weights together and then divide by the number of samples.\[ \bar{x} = \frac{117.6 + 109.5 + 111.6 + 109.2 + 119.1 + 110.8}{6} = \frac{677.8}{6} \approx 112.97 \text{ grams} \]
02

Perform a Hypothesis Test (Part a)

We need to determine if there is a reason to doubt that the population mean is at least 113 grams by performing a left-tailed hypothesis test.- Null hypothesis, \(H_0: \mu = 113\) grams- Alternative hypothesis, \(H_a: \mu < 113\) gramsCalculate the test statistic \(z\) using:\[ z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}} \]where \(\sigma\) is the population standard deviation we assume (we'll use the sample standard deviation in this step since \(\sigma\) isn't given), \(\bar{x} = 112.97\), and \(\mu_0 = 113\). Determine \(\sigma \) from sample data by first calculating the sample standard deviation \(s\).
03

Calculate Sample Standard Deviation

Find the sample standard deviation (\( s \)) using the following formula:\[ s = \sqrt{\frac{\sum{(x_i - \bar{x})^2}}{n-1}} \]where \(\bar{x} = 112.97\), \(n = 6\), and each \(x_i\) is one of the sample weights. Calculate:\[\sum{(x_i - \bar{x})^2} \approx 3.563^2 + 3.47^2 + 1.37^2 + 3.77^2 + 6.13^2 + 2.17^2 \approx 35.44 + 23.86 + 1.88 + 14.21 + 37.60 + 4.71 = 117.7 \]Then the sample standard deviation:\[s = \sqrt{\frac{117.7}{5}} \approx 4.858\]
04

Calculate Z-Value (Part a)

With the sample standard deviation calculated, compute the \( z \) value.\[ z = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{112.97 - 113}{4.858/\sqrt{6}} \approx \frac{-0.03}{1.983} \approx -0.015\]This \( z \)-value corresponds to the standard normal distribution. Compare it with critical value for \( \alpha = 0.05 \) of a left-tailed test. If the \( z \) value is less than the critical value of approximately -1.645, we reject \( H_0 \). In this case, we fail to reject \(H_0\).
05

Power of the Test under Hypothetical Conditions (Part b)

For part b, calculate the probability of rejecting \( H_0 \) when the true mean is 110 grams. Use the normal distribution's properties.The test statistic under the alternative hypothesis with true mean \( 110 \) is computed by resolving:\[ z' = \frac{\mu_0 - \mu_a}{\sigma/\sqrt{n}} \]where \( \mu_a = 110 \), \( \mu_0 = 113 \), \( \sigma = 4 \), and \( n = 6 \).\[ z' = \frac{110 - 113}{4/\sqrt{6}} = \frac{-3}{1.633} \approx -1.836 \]Find the power as the right tail beyond \(z'\) at \(z = -1.645\) using \(1 - \text{cumulative distribution function at } z'\).
06

Determine Required Sample Size (Part c)

To achieve a power of at least 0.95, find the necessary sample size.Calculate using:\[ n = \left( \frac{(Z_{1-\alpha} + Z_{1-\beta}) \sigma}{\mu_0 - \mu_a} \right)^2 \]where \(Z_{1-\alpha} = 1.645\) (critical value for 0.05 significance level), \(Z_{1-\beta} = 1.645\) (for power of 0.95), \(\sigma = 4\), and mean difference is 3.\[ n = \left( \frac{(1.645 + 1.645) \cdot 4}{3} \right)^2 = \left( \frac{3.29 \times 4}{3} \right)^2 = (4.387)^2 \approx 19.25 \]Thus, the required sample size is 20 (rounding up).

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a method used to determine whether a specific claim or hypothesis about a data set is true. In our exercise, we are interested in knowing if the weight of bagels is less than the claimed 113 grams. The process involves two main hypotheses:
  • The null hypothesis (\(H_0\)) which states that there is no effect or difference, here: \(\mu = 113\) grams.
  • The alternative hypothesis (\(H_a\)) suggesting a deviation from the null, here: \(\mu < 113\) grams.
Hypothesis testing involves calculating a test statistic, which allows us to measure how far the sample statistic diverges from the null hypothesis.
We then compare this statistic to a critical value, based on our significance level (\(\alpha\)), to decide whether to accept or reject the null hypothesis. If the test statistic is more extreme than the critical value, we reject the null hypothesis.
Sample Mean
The sample mean is the average of a set of data. In statistics, it provides an estimate of the overall population mean, which we are assessing.
For the bagel weights, the sample mean (\(\bar{x}\)) was calculated using:
\[ \bar{x} = \frac{\text{Sum of all sample weights}}{\text{Number of samples}} \]
Using our data, the sample mean is \(112.97\) grams. This mean is essential for comparing against the claimed mean of \(113\) grams in the hypothesis test.
The smaller the sample mean deviates from the population mean, the more evidence there is to support the claim that the sample reflects the population accurately.
Standard Deviation
The standard deviation is a measure of how spread out the values in a data set are around the mean. It provides insight into the variability within our bagel weight samples.
  • The formula for calculating the sample standard deviation is:
    \[ s = \sqrt{\frac{\sum{(x_i - \bar{x})^2}}{n-1}} \]
    where each \(x_i\) is an individual sample weight, \(\bar{x}\) is the sample mean, and \(n\) is the sample size.
In our case, the sample standard deviation turned out to be approximately \(4.858\) grams. A high standard deviation indicates that the data points are spread out over a larger range of values, whereas a low standard deviation signals that they are generally close to the sample mean.
For hypothesis testing, the standard deviation is crucial because it helps calculate the test statistic.
Power of a Test
The power of a test is the probability that it will correctly reject a false null hypothesis. In simpler terms, it measures the test's ability to detect an effect when it actually exists.
The power depends on several factors, including the sample size, significance level (\(\alpha\)), and the true effect size. Increasing the sample size generally increases the power, improving the test's accuracy.
  • In our bagel example, under assumption part b, when the mean is actually \(110\) grams contrary to the assumed \(113\), the test has a certain power calculated using the formula:
    \[ z' = \frac{\mu_0 - \mu_a}{\sigma/\sqrt{n}} \]
  • The power was found by checking the probability under the normal curve beyond a specific value matching the test statistic in alternative setup.
The exercise asked for additional observations to achieve a power of \(.95\), indicating a requirement for at least \(20\) observations. This ensures the test is robust enough to detect a true shift from the null hypothesis.

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Most popular questions from this chapter

A new design for the braking system on a certain type of car has been proposed. For the current system, the true average braking distance at 40 mph under specified conditions is known to be \(120 \mathrm{ft}\). It is proposed that the new design be implemented only if sample data strongly indicates a reduction in true average braking distance for the new design. a. Define the parameter of interest and state the relevant hypotheses. b. Suppose braking distance for the new system is normally distributed with \(\sigma=10\). Let \(\bar{X}\) denote the sample average braking distance for a random sample of 36 observations. Which of the following rejection regions is appropriate: \(R_{1}=\\{\bar{x}: \bar{x} \geq 124.80\\}, R_{2}=\) \(\\{\bar{x}: \bar{x} \leq 115.20\\}, R_{3}=\\{\bar{x}:\) either \(\bar{x} \geq 125.13\) or \(\bar{x} \leq 114.87\\} ?\) c. What is the significance level for the appropriate region of part (b)? How would you change the region to obtain a test with \(\alpha=.001\) ? d. What is the probability that the new design is not implemented when its true average braking distance is actually \(115 \mathrm{ft}\) and the appropriate region from part (b) is used? e. Let \(Z=(\bar{X}-120) /(\sigma / \sqrt{n})\). What is the significance level for the rejection region \(\\{z\) : \(z \leq-2.33\\}\) ? For the region \(\\{z: z \leq-2.88\\}\) ?

A manufacturer of nickel-hydrogen batteries randomly selects 100 nickel plates for test cells, cycles them a specified number of times, and determines that 14 of the plates have blistered. a. Does this provide compelling evidence for concluding that more than \(10 \%\) of all plates blister under such circumstances? State and test the appropriate hypotheses using a significance level of .05. In reaching your conclusion, what type of error might you have committed? b. If it is really the case that \(15 \%\) of all plates blister under these circumstances and a sample size of 100 is used, how likely is it that the null hypothesis of part (a) will not be rejected by the level 05 test? Answer this question for a sample size of 200 . c. How many plates would have to be tested to have \(\beta(.15)=.10\) for the test of part (a)?

When \(X_{1}, X_{2}, \ldots, X_{n}\) are independent Poisson variables, each with parameter \(\lambda\), and \(n\) is large, the sample mean \(\bar{X}\) has approximately a normal distribution with \(\mu=E(\bar{X})=\lambda\) and \(\sigma^{2}=V(\bar{X})=\) \(\lambda / n\). This implies that $$ Z=\frac{\bar{X}-\lambda}{\sqrt{\lambda / n}} $$ has approximately a standard normal distribution. For testing \(H_{0}: \lambda=\lambda_{0}\), we can replace \(\lambda\) by \(\lambda_{0}\) in the equation for \(Z\) to obtain a test statistic. This statistic is actually preferred to the large-sample statistic with denominator \(S / \sqrt{n}\) (when the \(X_{i}\) 's are Poisson) because it is tailored explicitly to the Poisson assumption. If the number of requests for consulting received by a certain statistician during a 5 -day work week has a Poisson distribution and the total number of consulting requests during a 36 -week period is 160 , does this suggest that the true average number of weekly requests exceeds 4.0? Test using \(\alpha=.02\).

Chapter 8 presented a CI for the variance \(\sigma^{2}\) of a normal population distribution. The key result there was that the rv \(\chi^{2}=(n-1) S^{2} / \sigma^{2}\) has a chi-squared distribution with \(n-1\) df. Consider the null hypothesis \(H_{0}: \sigma^{2}=\sigma_{0}^{2}\) (equivalently, \(\sigma=\sigma_{0}\) ). Then when \(H_{0}\) is true, the test statistic \(\chi^{2}=(n-1) S^{2} / \sigma_{0}^{2}\) has a chi-squared distribution with \(n-1\) df. If the relevant alternative is \(H_{\mathrm{a}}: \sigma^{2}>\sigma_{0}^{2}\), rejecting \(H_{0}\) if \((n-1) S^{2} / \sigma_{0}^{2} \geq\) \(\chi_{\alpha, n-1}^{2}\) gives a test with significance level \(\alpha\). To ensure reasonably uniform characteristics for a particular application, it is desired that the true standard deviation of the softening point of a certain type of petroleum pitch be at most \(.50^{\circ} \mathrm{C}\). The softening points of ten different specimens were determined, yielding a sample standard deviation of \(.58^{\circ} \mathrm{C}\). Does this strongly contradict the uniformity specification? Test the appropriate hypotheses using \(\alpha=.01\).

Let the test statistic \(Z\) have a standard normal distribution when \(H_{0}\) is true. Give the significance level for each of the following situations: a. \(H_{\mathrm{a}}: \mu>\mu_{0}\), rejection region \(z \geq 1.88\) b. \(H_{\mathrm{a}}: \mu<\mu_{0}\), rejection region \(z \leq-2.75\) c. \(H_{\mathrm{a}}: \mu \neq \mu_{0}\), rejection region \(z \geq 2.88\) or \(z \leq\) \(-2.88\)
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