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91Ó°ÊÓ

A candy company taste-tested two chocolate bars, one with almonds and one without almonds. A panel of testers rated the bars on a scale of 0 to \(5,\) with 5 indicating the highest taste rating. Assume the population standard deviations are equal. At the .05 significance level, do the ratings show a difference between chocolate bars with or without almonds? $$ \begin{array}{|ccc|} \hline \text { With Almonds } & \text { Without Almonds } \\ \hline 3 & 0 \\ 1 & 4 \\ 2 & 4 \\ 3 & 3 \\ 1 & 4 \\ 1 & \\ 2 & \\ \hline \end{array} $$

Short Answer

Expert verified
The ratings show a significant difference between chocolate bars with and without almonds.

Step by step solution

01

State the Hypotheses

We begin by stating the null and alternative hypotheses. The null hypothesis \( H_0 \) is that there is no difference between the mean ratings of the chocolate with almonds and without almonds. The alternative hypothesis \( H_a \) is that there is a difference. Mathematically, \( H_0: \mu_1 = \mu_2 \) and \( H_a: \mu_1 eq \mu_2 \).
02

Calculate Sample Means and Variances

Calculate the mean and variance for each sample. For 'With Almonds': \( \bar{x}_1 = \frac{3+1+2+3+1+1+2}{7} = \frac{13}{7} \approx 1.86 \) and for 'Without Almonds': \( \bar{x}_2 = \frac{0+4+4+3+4}{5} = \frac{15}{5} = 3 \). Use these sample means to find variances.
03

Calculate the Test Statistic

Since the population standard deviations are assumed equal, use the pooled variance. First find the pooled variance \( s_p^2 \): \[ s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} \] Here, \( s_1^2 = 0.8 \) and \( s_2^2 = 2.5 \). Then, the test statistic \( t \) is computed using \[ t = \frac{\bar{x}_1 - \bar{x}_2}{s_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \].
04

Determine the Critical Value and Decision Rule

With \( \alpha = 0.05 \) and degrees of freedom \( df = n_1 + n_2 - 2 = 10 \), find the critical \( t \)-value from the \( t \)-distribution table. For a two-tailed test, it's approximately \( \pm 2.228 \). Our decision rule is to reject \( H_0 \) if \( t \) is outside \([-2.228, 2.228]\).
05

Compare Test Statistic to the Critical Value

Substitute the values we calculated to find the test statistic, and then compare it to the critical value range. If the test statistic falls outside, we reject \( H_0 \).
06

Conclusion

Based on the comparison, if \( H_0 \) is rejected, it suggests that there is a significant difference in ratings between the two types of chocolate bars. If not, we do not have enough evidence to suggest a difference.

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Key Concepts

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

Null Hypothesis
The null hypothesis is a fundamental part of hypothesis testing. It's a statement made about the population that says there is no effect or no difference. In our case, the null hypothesis is that there is no difference in the mean taste ratings between chocolate bars with almonds and those without. Simply, this means we assume any observed difference in ratings is due to random chance and not a true difference.
Mathematically, we express this as:
  • \( H_0: \mu_1 = \mu_2 \)
Where \( \mu_1 \) and \( \mu_2 \) are the population means for the two chocolate types. If our analysis shows that \( H_0 \) cannot be rejected, it means our assumption of no difference holds true. But remember, not rejecting \( H_0 \) does not confirm it—it just suggests insufficient evidence to prove otherwise.
Alternative Hypothesis
The alternative hypothesis offers a statement that contradicts the null hypothesis. It is what you attempt to prove with your test. In our chocolate bar example, the alternative hypothesis claims there is a difference in mean taste ratings between chocolates with and without almonds.

Mathematically, we signify this as:
  • \( H_a: \mu_1 eq \mu_2 \)
Here, \( eq \) means 'not equal'. The focus is often on the alternative hypothesis because it signifies an effect or difference we're testing for. If the test data provides enough evidence to favor \( H_a \), we'll reject the null hypothesis in favor of \( H_a \). Understanding these hypotheses is key to deciphering the results of significance tests.
Pooled Variance
Pooled variance is used when we believe two populations have equal variances. In our candy tasting example, this assumption allows us to pool the variances from both samples into a single estimate. Pooled variance combines data from both groups to get a single, more reliable estimate of variance. Here's how it's calculated: \[ s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} \] Where:
  • \( s_1^2 \) and \( s_2^2 \) are the sample variances
  • \( n_1 \) and \( n_2 \) are sample sizes of the two groups
Pooled variance helps in computing the test statistic, which is crucial for hypothesis testing. By combining both variances, we achieve a more accurate estimate that influences the precision of our test results.
Test Statistic
The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It measures how much the sample data diverges from the null hypothesis. The formula varies depending on the specific test. In the scenario of comparing two means with equal variances, like the chocolate bars: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{s_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \] This formula compares the mean ratings from the two samples. Here:
  • \( \bar{x}_1 \) and \( \bar{x}_2 \) are sample means
  • \( s_p \) is the pooled standard deviation
  • \( n_1 \) and \( n_2 \) are sample sizes
The calculated value indicates how extreme the sample mean is, assuming the null hypothesis is true. A test statistic falling outside the critical value range points towards rejecting the null hypothesis.
Significance Level
The significance level, often denoted as \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true. It's a threshold set by the researcher which specifies the cutoff for determining whether a result is statistically significant. Common values for \( \alpha \) are 0.05, 0.01, and 0.10. In this exercise, we use \( 0.05 \).

Choosing \( \alpha = 0.05 \) means we're willing to accept a 5% chance of incorrectly rejecting the null hypothesis. It sets the critical values in hypothesis testing which determine our decision rule.
If our test statistic is more extreme than the critical value from the \( t \)-distribution at \( \alpha = 0.05 \), we reject the null hypothesis. Making sure you understand \( \alpha \) helps determine if the results are by random chance or they provide significant evidence.

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Most popular questions from this chapter

The management of Discount Furniture, a chain of discount furniture stores in the Northeast, designed an incentive plan for salespeople. To evaluate this innovative plan, 12 salespeople were selected at random, and their weekly incomes before and after the plan were recorded. $$ \begin{array}{|lrr|} \hline \text { Salesperson } & \text { Before } & \text { After } \\ \hline \text { Sid Mahone } & \$ 320 & \$ 340 \\ \text { Carol Quick } & 290 & 285 \\ \text { Tom Jackson } & 421 & 475 \\ \text { Andy Jones } & 510 & 510 \\ \text { Jean Sloan } & 210 & 210 \\ \text { Jack Walker } & 402 & 500 \\ \text { Peg Mancuso } & 625 & 631 \\ \text { Anita Loma } & 560 & 560 \\ \text { John Cuso } & 360 & 365 \\ \text { Carl Utz } & 431 & 431 \\ \text { A. S. Kushner } & 506 & 525 \\ \text { Fern Lawton } & 505 & 619 \\ \hline \end{array} $$ Was there a significant increase in the typical salesperson's weekly income due to the innovative incentive plan? Use the .05 significance level. Estimate the \(p\) -value, and interpret it.

Gibbs Baby Food Company wishes to compare the weight gain of infants using its brand versus its competitor's. A sample of 40 babies using the Gibbs products revealed a mean weight gain of 7.6 pounds in the first three months after birth. For the Gibbs brand, the population standard deviation of the sample is 2.3 pounds. A sample of 55 babies using the competitor's brand revealed a mean increase in weight of 8.1 pounds. The population standard deviation is 2.9 pounds. At the .05 significance level, can we conclude that babies using the Gibbs brand gained less weight? Compute the \(p\) -value and interpret it.

The amount of income spent on housing is an important component of the cost of living. The total costs of housing for homeowners might include mortgage payments, property taxes, and utility costs (water, heat, electricity). An economist selected a sample of 20 homeowners in New England and then calculated these total housing costs as a percent of monthly income, 5 years ago and now. The information is reported below. Is it reasonable to conclude the percent is less now than 5 years ago? $$ \begin{array}{|llllll|} \hline \text { Homeowner } & \text { Five Years Ago } & \text { Now } & \text { Homeowner } & \text { Five Years Ago } & \text { Now } \\ \hline \text { Holt } & 17 \% & 10 \% & \text { Lozier } & 35 \% & 32 \% \\ \text { Pierse } & 20 & 39 & \text { Cieslinski } & 16 & 32 \\ \text { Merenick } & 29 & 37 & \text { Rowatti } & 23 & 21 \\ \text { Lanoue } & 43 & 27 & \text { Koppel } & 33 & 12 \\ \text { Fagan } & 36 & 12 & \text { Rumsey } & 44 & 40 \\ \text { Bobko } & 43 & 41 & \text { McGinnis } & 44 & 42 \\ \text { Kippert } & 45 & 24 & \text { Pierce } & 28 & 22 \\ \text { San Roman } & 19 & 26 & \text { Roll } & 29 & 19 \\ \text { Kurimsky } & 49 & 28 & \text { Lang } & 39 & 35 \\ \text { Davison } & 49 & 26 & \text { Miller } & 22 & 12 \\ \hline \end{array} $$

Fry Brothers Heating and Air Conditioning Inc. employs Larry Clark and George Murnen to make service calls to repair furnaces and air-conditioning units in homes. Tom Fry, the owner, would like to know whether there is a difference in the mean number of service calls they make per day. A random sample of 40 days last year showed that Larry Clark made an average of 4.77 calls per day. For a sample of 50 days, George Murnen made an average of 5.02 calls per day. Assume the population standard deviation is 1.05 calls per day for Larry Clark and 1.23 calls per day for George Murnen. At the .05 significance level, is there a difference in the mean number of calls per day between the two employees? What is the \(p\) -value?

A recent study compared the time spent together by single- and dual-earner couples. According to the records kept by the wives during the study, the mean amount of time spent together watching television among the single-earner couples was 61 minutes per day, with a standard deviation of 15.5 minutes. For the dual-earner couples, the mean number of minutes spent watching television was 48.4 minutes, with a standard deviation of 18.1 minutes. At the .01 significance level, can we conclude that the single-earner couples on average spend more time watching television together? There were 15 single-earner and 12 dual-earner couples studied.

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