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Chapter 9: Problem 37

Taste Testing In a head-to-head taste test of store-brand foods versus national brands, Consumer Reports found that it was hard to find a taste difference in the two. \({ }^{10}\) If the national brand is indeed better than the store brand, it should be judged as better more than \(50 \%\) of the time. a. State the null and alternative hypothesis to be tested. Is this a one- or a two-tailed test? b. Suppose that, of the 35 food categories used for the taste test, the national brand was found to be better than the store brand in eight of the taste comparisons. Use this information to test the hypothesis in part a. Use \(\alpha=.01\). What practical conclusions can you draw from the results?

Short Answer

Expert verified
Answer: No, the national brand is not preferred more than 50% of the time when compared to the store brand in terms of taste.

Step by step solution

01

Define null and alternative hypotheses and determine the type of test

Given that the national brand should be judged as better more than \(50 \%\) of the time if it is indeed better than the store brand, we can define our null and alternative hypothesis as follows: Null Hypothesis \(H_0\): The national brand and store brand are equally preferred (judged better \(50 \%\) of the time). Alternative Hypothesis \(H_1\): The national brand is preferred more than \(50 \%\) of the time (judged better more than half the time). This is a one-tailed test since we are only interested in whether the national brand is judged better more than \(50 \%\) of the time.
02

Calculate the test statistic

Given that the national brand was found to be better in eight out of the 35 food categories, the sample proportion is: $$ \hat{p} = \frac{8}{35} = 0.2286 $$ Now we need to calculate the standard error for the sample proportion: $$ SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5(1-0.5)}{35}} = 0.0802 $$ Next, we will calculate the test statistic: $$ z =\frac{\hat{p} - p}{SE} = \frac{0.2286 - 0.5}{0.0802} = -3.383 $$
03

Find the p-value and conduct the hypothesis test

The significance level for this test is \(\alpha = 0.01\). Since this is a one-tailed test, we will look up the p-value corresponding to the z-value in a z-table. The p-value for \(z = -3.383\) is approximately \(0.0004\). Now we compare the p-value to the significance level: $$ 0.0004 < 0.01 $$ Since the p-value is less than the significance level, we reject the null hypothesis in favor of the alternative hypothesis.
04

Draw practical conclusions

Since we rejected the null hypothesis, we can conclude that the national brand is not preferred more than \(50 \%\) of the time. In fact, it was only judged better in eight out of the 35 food categories. This suggests that, in terms of taste, the national brand may not hold a significant advantage over the store brand. Thus, consumers might consider purchasing store-brand foods that offer similar quality at potentially lower prices.

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Key Concepts

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

null hypothesis
In statistical hypothesis testing, the null hypothesis, often denoted as \(H_0\), represents a default position that there is no effect or no difference. It's like saying "nothing is happening, or nothing has changed." For this exercise, the null hypothesis states that the national brand and the store brand are equally preferred, with each being judged better 50% of the time.
It's important to clearly define the null hypothesis because it sets the stage for comparison. When we perform tests, we're essentially trying to find evidence against this default position. If, after performing the test, we find significant evidence that suggests otherwise, we may reject the null hypothesis.
Here, the null hypothesis is formulated as follows:
  • \(H_0\): The national brand and store brand are equally preferred (judged better 50% of the time).
Understanding the null hypothesis allows us to measure the strength of evidence within the context given and decides whether any observed difference or outcome is due to random chance.
alternative hypothesis
The alternative hypothesis, denoted as \(H_1\), is the statement we consider if there is enough evidence to reject the null hypothesis. In essence, it is what we might believe if the null hypothesis is found to be unlikely given our data.
In our taste test scenario, the alternative hypothesis suggests that the national brand is preferred more than 50% of the time, indicating a difference between the two brands.
The alternative hypothesis is stated as:
  • \(H_1\): The national brand is preferred more than 50% of the time (judged better more than half the time).
It's important to define whether the test is one-tailed or two-tailed. Here, since we are only interested in whether the national brand is preferred more (not just different) than 50%, the test is one-tailed. The alternative hypothesis is critical because it represents what the researchers want to prove and is central to any conclusions drawn from the hypothesis testing.
significance level
The significance level, often represented by \(\alpha\), is a threshold of probability which determines when we reject the null hypothesis. It is one of the key components of hypothesis testing that helps us decide how strong our evidence must be to convince us that the null hypothesis isn't true.
Commonly used significance levels are 0.05, 0.01, and 0.10, where a lower significance level requires stronger evidence to reject the null hypothesis. In this exercise, a significance level of \(\alpha = 0.01\) is used, which indicates a very rigorous test aiming to minimize false positives.
The choice of significance level depends on the field of study and the consequences of making errors. For instance:
  • A significance level of 0.01 means there is a 1% risk of concluding that a difference or effect exists when there is none (Type I error).
Understanding the significance level allows researchers and analysts to make informed conclusions about data and whether the observed effect is likely to be true or just a result of chance.

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Most popular questions from this chapter

Sweet Potato Whitefly Suppose that \(10 \%\) of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in this area are randomly selected, and 25 are found to be infested with whitefly. a. Assuming that the experiment satisfies the conditions of the binomial experiment, do the data indicate that the proportion of infested fields is greater than expected? Use the \(p\) -value approach, and test using a \(5 \%\) significance level b. If the proportion of infested fields is found to be significantly greater than .10 , why is this of practical significance to the agronomist? What practical conclusions might she draw from the results?

Washing Machine Colors A manufacturer of automatic washers provides a particular model in one of three colors. Of the first 1000 washers sold, it is noted that 400 were of the first color. Can you conclude that more than one- third of all customers have a preference for the first color? a. Find the \(p\) -value for the test. b. If you plan to conduct your test using \(\alpha=.05\), what will be your test conclusions?

Flextime A company was contemplating the installation of a flextime schedule in which a worker schedules his or her work hours or compresses work weeks. The company estimates that it needs a minimum mean of 7 hours per day per assembly worker in order to operate effectively. Each of a random sample of 80 of the company's assemblers was asked to submit a tentative flextime schedule. If the mean number of hours per day for Monday was 6.7 hours and the standard deviation was 2.7 hours, do the data provide sufficient evidence to indicate that the mean number of hours worked per day on Mondays, for all of the company's assemblers, will be less than 7 hours? Test using \(\alpha=.05 .\)

What's Normal? What is normal, when it comes to people's body temperatures? A random sample of 130 human body temperatures, provided by Allen Shoemaker \(^{3}\) in the Journal of Statistical Education, had a mean of \(98.25^{\circ}\) and a standard deviation of \(0.73^{\circ} .\) Does the data indicate that the average body temperature for healthy humans is different from \(98.6^{\circ},\) the usual average temperature cited by physicians and others? Test using both methods given in this section. a. Use the \(p\) -value approach with \(\alpha=.05\). b. Use the critical value approach with \(\alpha=.05\). c. Compare the conclusions from parts a and b. Are they the same? d. The 98.6 standard was derived by a German doctor in 1868 , who claimed to have recorded 1 million temperatures in the course of his research. \({ }^{4}\) What conclusions can you draw about his research in light of your conclusions in parts a and b?

Noise and Stress In Exercise \(8.52,\) you compared the effect of stress in the form of noise on the ability to perform a simple task. Seventy subjects were divided into two groups; the first group of 30 subjects acted as a control, while the second group of 40 was the experimental group. Although each subject performed the task, the experimental group subjects had to perform the task while loud rock music was played. The time to finish the task was recorded for each subject and the following summary was obtained: $$\begin{array}{lll} & \text { Control } & \text { Experimental } \\\\\hline n & 30 & 40 \\\\\bar{x} & 15 \text { minutes } & 23 \text { minutes } \\\s & 4 \text {minutes } & 10 \text { minutes }\end{array}$$
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