/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 A dairy farm uses the somatic ce... [FREE SOLUTION] | 91Ó°ÊÓ

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A dairy farm uses the somatic cell count (SCC) report on the milk it provides to a processor as one way to monitor the health of its herd. The mean SCC from five samples of raw milk was 250,000 cells per milliliter with standard deviation 37,500 cell/ml. Test whether these data provide sufficient evidence, at the \(10 \%\) level of significance, to conclude that the mean SCC of all milk produced at the dairy exceeds that in the previous report, 210,250 cell/ml. Assume a normal distribution of SCC.

Short Answer

Expert verified
Reject \( H_0 \); the mean SCC exceeds 210,250 cells/ml.

Step by step solution

01

Understanding the Hypotheses

First, we establish our null hypothesis \( H_0 \) and alternative hypothesis \( H_a \). Here, \( H_0: \mu = 210,250 \) cells/ml (the mean SCC is the same as before) and \( H_a: \mu > 210,250 \) cells/ml (the mean SCC is greater than before). We are conducting a one-tailed test since we are only interested in checking for an increase.
02

Collecting Sample Data

From the problem, we have a sample mean \( \bar{x} = 250,000 \) cells/ml, a sample standard deviation \( s = 37,500 \) cells/ml, and a sample size \( n = 5 \).
03

Calculate the Test Statistic

Use the formula for a t-test statistic: \( t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \). Plugging in the numbers gives \( t = \frac{250,000 - 210,250}{37,500/\sqrt{5}} \).
04

Computing the Test Statistic

Calculate the exact value of the test statistic: \( t = \frac{39,750}{16,746.63} \approx 2.37 \).
05

Determine the Critical Value

Since the level of significance is \( \alpha = 0.10 \) and it's a one-tailed test, we look up the t-distribution table for \( n-1 = 4 \) degrees of freedom. The critical value \( t_{0.10, 4} \approx 1.533 \).
06

Decision Rule

If the calculated \( t \)-statistic is greater than the critical value from the t-table, we reject the null hypothesis. Here, \( t = 2.37 \) is greater than \( 1.533 \).
07

Conclusion

Since \( t_{calculated} > t_{critical} \), we reject the null hypothesis \( H_0 \). There is sufficient evidence at the \( 10\% \) significance level to conclude that the mean SCC exceeds 210,250 cells/ml.

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Key Concepts

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

t-Distribution
When working with small sample sizes, like in this problem where only five milk samples are analyzed, researchers often rely on the t-distribution, particularly the Student’s t-distribution. This is because it accounts for the added variability that can occur in smaller samples when compared to larger ones. The t-distribution is similar to the normal distribution but has fatter tails. This means there is a greater likelihood of obtaining values far from the mean. This characteristic is particularly useful when we're working with small datasets, ensuring that our statistical tests remain accurate, and giving us a reliable method for estimating population parameters.
In the context of this exercise, critical t-values are necessary to determine whether to reject the null hypothesis. By knowing the degrees of freedom (in this case, the sample size minus one, n-1), we can look up these values in the t-distribution table. For the sample of 5, our degrees of freedom are 4. Understanding the role of the t-distribution is crucial for interpreting results in hypothesis testing, especially when sample sizes are small.
One-Tailed Test
In hypothesis testing, a one-tailed test is used when our research question or alternative hypothesis is directional. In this problem, the goal is to check if the mean SCC in the current sample exceeds the mean from a prior report. This directional hypothesis leads us to use a one-tailed test focusing on the possible increase. There are two types of one-tailed tests: left-tailed and right-tailed. Our test is a right-tailed one because we are only interested in whether the mean is greater than before.
Using a one-tailed test gives us more power to detect an effect in one direction at the expense of detecting an effect in the opposite direction. Thus, it is indispensable when the research focuses on a specific direction of change. When looking at the critical t-value in a one-tailed test, we only consider one end of the distribution, in this case, the upper tail. This makes determining conclusions for problems such as this one more precise and relevant.
Level of Significance
The level of significance, often denoted as \( \alpha \), is a critical component of hypothesis testing. It represents the threshold at which we agree to reject the null hypothesis. In this exercise, the level of significance is set at 10%, or \( \alpha = 0.10 \). This implies there is a 10% risk of concluding that the mean SCC is greater than 210,250 cells/ml when it might not be.
Choosing \( \alpha \) depends on the research context and the balance between Type I and Type II errors. Smaller \( \alpha \) values, like 0.01 or 0.05, reduce the likelihood of incorrectly rejecting the null hypothesis but require stronger evidence to do so. A higher level, like 0.10, indicates a more lenient criterion for rejecting the null hypothesis, reserved for situations when missing a potential effect (Type II error) is more critical than a false positive (Type I error). Determining and understanding this level help shape the decision-making process in statistical testing.
Null Hypothesis
The null hypothesis \( H_0 \) is a foundational concept in hypothesis testing. It represents the default statement or position that there is no effect or no change. For this problem, \( H_0: \mu = 210,250 \) cells/ml suggests that the mean SCC is the same as reported previously. The null hypothesis is what researchers initially set out to test against, using statistical data.
In hypothesis tests, researchers either reject the null hypothesis, indicating sufficient evidence for the alternative hypothesis, or fail to reject it, meaning the evidence is insufficient for the alternative. It is crucial to understand that failing to reject the null does not prove it true, it simply indicates a lack of evidence against it. Every decision rule in hypothesis testing revolves around whether or not the gathered evidence crosses the threshold set by the level of significance to reject \( H_0 \). Clearly defining the null hypothesis is necessary for setting up and carrying out effective testing, as it lays the groundwork for interpretation and decisions.

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Most popular questions from this chapter

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: \(\mu \neq 0.12 @ \alpha=0.001\).

The recommended daily allowance of iron for females aged \(19-50\) is \(18 \mathrm{mg} /\) day. A careful measurement of the daily iron intake of 15 women yielded a mean daily intake of \(16.2 \mathrm{mg}\) with sample standard deviation 4.7 \(\mathrm{mg} .\) a. Assuming that daily iron intake in women is normally distributed, perform the test that the actual mean daily intake for all women is different from \(18 \mathrm{mg} /\) day, at the \(10 \%\) level of significance. b. The sample mean is less than 18, suggesting that the actual population mean is less than 18 \(\mathrm{mg} /\) day. Perform this test, also at the \(10 \%\) level of significance. (The computation of the test statistic done in part (a) still applies here.)

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: \(\mu \neq-375 @ \alpha=0.01, \sigma=18.5, n=31, x-=-388, s=18.0\)

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.)

State the null and alternative hypotheses for each of the following situations. (That is, identify the correct number \(\mu_{0}\) and write \(H_{0} \cdot \mu=\mu_{0}\) and the appropriate analogous expression for \(H_{a}\).) a. The average July temperature in a region historically has been \(74.5^{\circ} \mathrm{F}\). Perhaps it is higher now. b. The average weight of a female airline passenger with luggage was 145 pounds ten years ago. The FAA believes it to be higher now. c. The average stipend for doctoral students in a particular discipline at a state university is \(\$ 14,756\). The department chairman believes that the national average is higher. d. The average room rate in hotels in a certain region is \(\$ 82.53 .\) A travel agent believes that the average in a particular resort area is different. e. The average farm size in a predominately rural state was 69.4 acres. The secretary of agriculture of that state asserts that it is less today.

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