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Lauren, a brunette, was tired of hearing, "Blondes have more fun." She set out to "prove" that "brunettes are more intelligent." Lauren randomly (as best she could) selected 40 blondes and 40 brunettes at her high school.$$\begin{array}{llll}\hline \text { Blondes } & n_{\beta}=40 & \bar{x}_{\beta i}=88.375 & s_{\beta i}=6.134\\\\\text { Brunettes } & n_{\mathrm{S}}=40 & \bar{x}_{A r}=87.600 & s_{B r}=6.640\\\\\hline\end{array}$$ Upon seeing the sample results, does Lauren have support for her claim that "brunettes are more intelligent than blondes"? Explain. What could Lauren say about blondes' and brunettes' intelligence?

Short Answer

Expert verified
The conclusion will be based on the t-test result. If the calculated t is larger than the critical t, we may conclude that brunettes on average are more intelligent than blondes with statistical significance. In any case, it is important to note that this test does not apply to individuals but averages of groups, and intelligence can vary greatly within each group.

Step by step solution

01

Identify the hypothesis

The hypotheses for this problem would be set up as follows: \n Null Hypothesis (\(H_0\)): The average intelligence scores of both groups are equal (\(µ_{brunettes}= µ_{blondes}\)).\n Alternative Hypothesis (\(H_a\)): The average intelligence score of the brunettes is larger than that of the blondes (\(µ_{brunettes}> µ_{blondes}\)).\n It's a one-tailed test because the focus is only on whether brunettes are more intelligent than blondes.
02

Calculate the t statistic

Next, calculate the t statistic using the t formula for the test of difference of means.\n \( t = \frac{\bar{x}_1 - \bar{x}_2 - D_0}{\sqrt{\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2}}}\) \n Where:\n \(\bar{x}_1\) and \(\bar{x}_2\) = means of the two samples.\n \(s_1\) and \(s_2\) = standard deviations of the two samples.\n \(n_1\) and \(n_2\) = sizes of the two samples.\n \(D_0\) = hypothesized value of difference in means which is zero in this case.\n On substituting the given values in t-formula, we compute the t statistic.
03

Find the critical value and make decision

Once the t-statistic is calculated, find the critical value for the one-tailed test with a significance level (α), commonly 0.05. The degrees of freedom is \(n_{1} + n_{2} - 2\). Compare the calculated t value with the critical t value. If the calculated t is higher than the critical t, we reject the null hypothesis and conclude that brunettes have a higher average level of intelligence than blondes.
04

Interpret the result

The conclusion based on the test outcome determines whether or not the data supports Lauren's claim that brunettes are more intelligent. Furthermore, Lauren can comment on the range of scores and variation in them.

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Key Concepts

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

Null and Alternative Hypotheses
Creating a hypothesis is a critical part of conducting any statistical test. In hypothesis testing, we set up two potential statements about the data: the Null Hypothesis and the Alternative Hypothesis.
The **Null Hypothesis** (\(H_0\)) proposes that there is no effect or difference between groups. It serves as a baseline that suggests any observed effect is due to random sampling error rather than a true effect. In Lauren's study, the null hypothesis is that both blondes and brunettes have the same average intelligence scores, mathematically expressed as\( \mu_{brunettes} = \mu_{blondes} \).
On the other hand, the **Alternative Hypothesis** (\(H_a\)) suggests that there is a real difference or effect present. It is what you aim to support with your test. For Lauren's case, the alternative is that brunettes have higher average intelligence scores than blondes, expressed as\( \mu_{brunettes} > \mu_{blondes} \).
These hypotheses form the foundation of statistical decision-making, guiding the rest of the analysis to determine whether the observed data supports the null or the alternative.
One-tailed Test
When conducting a hypothesis test, you have to decide between a one-tailed and a two-tailed test. The choice depends on the direction of the hypothesis tested.
A **one-tailed test** is used when the research hypothesis predicts a specific direction of the difference. In Lauren's scenario, her claim that brunettes are more intelligent than blondes is directional, prompting the use of a one-tailed test.
With a one-tailed test, the entire alpha level (commonly 0.05) is allocated to testing the significance of the expected direction, increasing the sensitivity to detect an effect in that direction.
  • Focus: Whether brunettes have a higher average intelligence score than blondes.
  • Allocation: The significance level (alpha) is wholly devoted to one side of the distribution.
This choice increases test power for detecting a difference but sacrifices the ability to detect effects in the opposite direction.
t statistic calculation
The calculation of the t statistic is pivotal in hypothesis testing as it translates sample data into a measurement that can be compared against known statistical distributions. Here's how it works:
To calculate the t statistic, we use the formula:\(t = \frac{\bar{x}_1 - \bar{x}_2 - D_0}{\sqrt{\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2}}}\)where:
  • \(\bar{x}_1 \) and \(\bar{x}_2\) are the sample means for brunettes and blondes, respectively.
  • \(s_1\) and \(s_2\) are the standard deviations for the groups.
  • \(n_1\) and \(n_2\) represent sample sizes, which are equal here.
  • \(D_0\) is the difference in population means under the null hypothesis, which is zero.
This formula provides a standardized way to measure how far the sample statistic lies from the population parameter specified in the null hypothesis, considering the variability and sample sizes.
Once calculated, the t statistic helps decide if the observed difference (or lack thereof) is significant.
Significance Level
The significance level, often denoted by alpha (\( \alpha \)), is a threshold set before conducting the test that determines how extreme the data must be to reject the null hypothesis.
**Common choices for alpha include 0.05 or 0.01**, indicating a 5% or 1% chance of incorrectly rejecting the null hypothesis when it is actually true (Type I error). For a more stringent test, a smaller alpha such as 0.01 might be chosen.
In a hypothesis test, once you calculate the test statistic (such as a t value), you compare it with a critical value from a statistical distribution table. This critical value depends on the chosen alpha level and the degrees of freedom (which is calculated as the sum of the sample sizes minus two for two samples).
  • If your calculated statistic exceeds this critical value in a one-tailed test, you reject the null hypothesis.
  • Rejecting the null means there's statistical support for the alternative hypothesis.
Understanding the significance level in this way helps you appreciate the balance between false positives and confidence in the statistical conclusions.

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Most popular questions from this chapter

Are females as serious about golf as men are? If so, would the price of a driver for a man be the same as the price of a driver for a female? It was contended that women's drivers would be cheaper. Random samples of drivers were taken from the Golflink.com website. The prices were: $$\begin{array}{llllll} \text { Male } & & & & & \\\\\hline 149.99 & 299.99 & 49.99 & 499.99 & 167.97 & 299.99 \\\399.99 & 199.99 & 99.99 & 149.99 & & \\\\\hline \text { Female } & & & & & \\\\\hline 199.99 & 79.99 & 499.99 & 199.97 & 299.99 & 99.99\end{array}$$ At the 0.05 level of significance, is there sufficient evidence to support the contention that men's drivers are more expensive than women's drivers? Assume normality of golf driver prices.

a. Two independent samples, each of size \(3,\) are drawn from a normally distributed population. Find the probability that one of the sample variances is at least 19 times larger than the other one. b. Two independent samples, each of size \(6,\) are drawn from a normally distributed population. Find the probability that one of the sample variances is no more than 11 times larger than the other one.

At a large university, a mathematics placement exam is administered to all students. Samples of 36 male and 30 female students are randomly selected from this year's student body and the following scores recorded: $$\begin{array}{lllllllllll}\hline \text { Male } & 72 & 68 & 75 & 82 & 81 & 60 & 75 & 85 & 80 & 70 \\\& 71 & 84 & 68 & 85 & 82 & 80 & 54 & 81 & 86 & 79 \\\& 99 & 90 & 68 & 82 & 60 & 63 & 67 & 72 & 77 & 51 \\\& 61 & 71 & 81 & 74 & 79 & 76 & & & & \\\\\hline \text { Female } & 81 & 76 & 94 & 89 & 83 & 78 & 85 & 91 & 83 & 83 \\\& 84 & 80 & 84 & 88 & 77 & 74 & 63 & 69 & 80 & 82 \\\& 89 & 69 & 74 & 97 & 73 & 79 & 55 & 76 & 78 & 81 \\ \hline\end{array}$$ a. Describe each set of data with a histogram (use the same class intervals on both histograms), the mean, and the standard deviation. b. Construct \(95 \%\) confidence interval for the mean score for all male students. Do the same for all female students. c. Do the results found in part b show that the mean scores for males and females could be the same? Justify your answer. Be careful! d. Construct the \(95 \%\) confidence interval for the difference between the mean scores for male and female students. e. Do the results found in part d show that the mean scores for male and female students could be the same? Explain. f. Explain why the results in part b cannot be used to draw conclusions about the difference between the two means.

Determine the \(p\) -value for the following hypothesis tests for the difference between two means with population variances unknown. a. \(\quad H_{a}: \mu_{1}-\mu_{2}>0, n_{1}=6, n_{2}=10, t \star=1.3\) b. \(\quad H_{a}: \mu_{1}-\mu_{2}<0, n_{1}=16, n_{2}=9, t \star=-2.8\) c. \(\quad H_{a}: \mu_{1}-\mu_{2} \neq 0, n_{1}=26, n_{2}=16, t \star=1.8\) d. \(\quad H_{a}: \mu_{1}-\mu_{2} \neq 5, n_{1}=26, n_{2}=35, t *=-1.8\)

With all the media attention from the Dancing With the Stars program, local dance studios are seeing an upswing in the numbers of people interested in taking ballroom dancing lessons. Two samples of 15 students are to be judged before taking any lessons and then again after five lessons. The students belonging to the samples are to be randomly selected. a. How can data be collected if dependent samples are to be obtained? Explain in detail. b. How can data be collected if independent samples are to be obtained? Explain in detail.

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