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A candy company has developed two new chocolate-covered candy bars. Six randomly selected people all preferred candy bar I. Is this statistical evidence, at \(\alpha=0.05,\) that the general public will prefer candy bar I?

Short Answer

Expert verified
Yes, this is statistical evidence at significance level \(\alpha = 0.05\), that more than half of the general public will prefer candy bar I.

Step by step solution

01

Define the Null and Alternative Hypothesis

In this case, hypotheses will be formulated in terms of the population proportion \(p\) that prefers candy bar I. The null hypothesis is that half or less than half of the population will prefer candy bar I (\(H_0: p \leq 0.5\)). The alternative hypothesis is that more than half of the population will prefer candy bar I (\(H_a: p > 0.5\)).
02

Determine the Critical Value

Next is to determine the critical value corresponding to the \(\alpha=0.05\) level of significance. Z-statistic for a binomial proportion test under null hypothesis condition is calculated as follows: \(Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\) where \(Z\) is the z-statistic, \(\hat{p}\) is the sample proportion, \(p_0\) is hypothesized value of population proportion, \(n\) is the sample size. In this problem, \(\hat{p} = \frac{6}{6} = 1\) (all six people preferred candy bar I), \(p_0 = 0.5\) (if null hypothesis is true), and \(n = 6\) (the number of people in the sample). Substituting in the formula gives \(Z \approx 2.449\). The critical value for the level of significance \(0.05\) in a one-tailed test, Zα = 1.645.
03

Conduct the Test and Interpret the Result

Computed Z-value is greater than the critical value (2.449 > 1.645). Thus, we reject the null hypothesis in favor of the alternative hypothesis. This concludes that there is significant evidence to suggest that more than half of the general public will prefer candy bar I.

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Key Concepts

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

Null and Alternative Hypothesis
Understanding null and alternative hypotheses is the foundation of hypothesis testing. The null hypothesis (\(H_0\) ) represents the status quo or a position of skepticism that there is no effect or no difference.
For our candy company example, the null hypothesis states that half or fewer people will prefer candy bar I - formally written as
\[\begin{equation}H_0: p \<= 0.5.\end{equation}\]The alternative hypothesis (\(H_a\) or \(H_1\)), on the other hand, denotes what the researcher is seeking evidence for. It challenges the status quo.In this case, the alternative hypothesis avers that more than half of the population prefers candy bar I, expressed as\[\begin{equation}H_a: p > 0.5.\end{equation}\]It's important to note that the null hypothesis is assumed true until evidence suggests otherwise. The aim of hypothesis testing is to determine whether there is enough statistical evidence in the sample data to reject the null hypothesis in favor of the alternative. If the evidence suggests the contrary, the null hypothesis cannot be rejected.
Critical Value
The concept of a critical value is related to the level of significance of a test, usually denoted by \(\alpha\) . It serves as a threshold, which determines whether the observed test statistic is extreme enough to reject the null hypothesis.In hypothesis testing, we compare the test statistic to this critical value. If the test statistic falls within critical regions (beyond the critical value in the direction of the alternative hypothesis), it suggests that the null hypothesis is unlikely.For the given candy bar preference test at a 0.05 significance level (\(\alpha = 0.05\) ), the critical value is the point in the distribution that has 5% of the observations above it in a one-tailed test. For \(\alpha = 0.05\) in a one-tailed Z-distribution, the critical value is approximately 1.645.This means if our calculated Z-statistic is greater than 1.645, we have sufficient evidence at the 5% level of significance to reject the null hypothesis.
Z-statistic
The Z-statistic is a type of standard score that indicates how many standard deviations an element is from the mean. In hypothesis testing, the Z-statistic is used to measure the difference between the sample statistic and the null hypothesis. It's calculated using the following formula:\[\begin{equation}Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\end{equation}\]where \(\hat{p}\) is the sample proportion, \(p_0\) is the hypothesized value of population proportion under the null hypothesis, and \(n\) is the sample size.In our candy bar scenario, the Z-statistic was calculated to be 2.449 which is more about what happens when \(\alpha = 0.05\) for a one-tailed test. Since 2.449 exceeds the critical value of 1.645, the null hypothesis is rejected. This means we have found significant evidence to suggest that more than half of the public may indeed prefer candy bar I.

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Most popular questions from this chapter

Research Randomizer is a free service offered to students and researchers interested in conducting random assignment and random sampling. Although every effort has been made to develop a useful means of generating random numbers, Research Randomizer and its staff do not guarantee the quality or randomness of numbers they generate. Any use to which these numbers are put remains the sole responsibility of the user who generated them. a. Go to the website http://www.randomizer.org/ about.htm and generate one set of 20 random numbers from 1 to 9, where each number can repeat (select "No" for each number to be unique). (Use your computer, calculator, or Table 1 in the back, if you do not have a Web connection.) b. Test your set for randomness above and below the median value of \(5 .\) Use \(\alpha=0.05\). c. Conduct the test again with the same parameters. d. Test your new set for randomness. Use \(\alpha=0.05\) Did you get the same results? e. Solve part d using the standard normal distribution. Did you reach the same conclusion?

State the null hypothesis, \(H_{o}\), and the alternative hypothesis, \(H_{a},\) that would be used to test the following statements: a. There is no relationship between the two rankings. b. The two variables are unrelated. c. There is a positive correlation between the two variables. d. Refrigerator age has a decreasing effect on monetary value.

Determine the critical value that would be used to test the following hypotheses for experiments involving two independent samples, using the classical method: a. \(H_{o}:\) Average \((\mathrm{A})=\) Average \((\mathrm{B})\) \(H_{a}:\) Average \((\mathrm{A}) > \) Average \((\mathrm{B})\) with \(n_{\mathrm{A}}=18, n_{\mathrm{B}}=15,\) and \(\alpha=0.05\). b. \(H_{o}:\) The average score is the same for both groups; \(H_{a}:\) Group I average score is lower than group II; with \(n_{1}=78, n_{11}=45,\) and \(\alpha=0.05\).

State the null hypothesis, \(H_{o},\) and the alternative hypothesis, \(H_{a},\) that would be used to test the following statements: a. The students in the new reading program scored higher on the comprehension test than the students in the traditional reading programs. b. Men on the grapefruit diet lose more weight than men not on the grapefruit diet. c. There is no difference in growth between the use of the two fertilizers.

a. What parametric test procedure is comparable to the Mann-Whitney \(U\) test? b. What characteristic of the data used in a parametric test is not used in the Mann-Whitney \(U\) test?

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