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Chapter 10: Problem 52

A company sent seven of its employees to attend a course in building self- confidence. These employees were evaluated for their self-confidence before and after attending this course. The following table gives the scores (on a scale of 1 to 15,1 being the lowest and 15 being the highest score) of these employees before and after they attended the course. $$ \begin{array}{l|rrrrrrr} \hline \text { Before } & 8 & 5 & 4 & 9 & 6 & 9 & 5 \\ \hline \text { After } & 10 & 8 & 5 & 11 & 6 & 7 & 9 \\ \hline \end{array} $$ a. Construct a \(95 \%\) confidence interval for the mean \(\mu_{d}\) of the population paired differences, where a paired difference is equal to the score of an employee before attending the course minus the score of the same employee after attending the course b. Test at the \(1 \%\) significance level whether attending this course increases the mean score of employees.

Short Answer

Expert verified
The confidence interval gives a range of the scores that shows an improvement in self-confidence after attending the course and the hypothesis test confirms whether these improvements are statistically significant. The conclusion of this test and interval will vary depending on the calculated values from the specific score data.

Step by step solution

01

Calculating the Differences

For each employee, calculate the difference in the scores before and after attending the course. This can be done by subtracting the 'after' score from the 'before' score
02

Calculate the Mean and Standard Deviation of the Differences

Find the mean (average) and standard deviation of these differences. Let's denote them as \( \bar{d} \) and \( s_d \) respectively
03

Construct the Confidence Interval

After calculating the mean and standard deviation, we can construct the 95% confidence interval for the mean of the paired differences \(\mu_d\). Convert 95% confidence level into z-score, which is around 1.96 in this case. Use the following formula: \( CI = \bar{d} \pm Z*\frac{s_d}{\sqrt{n}} \), where n is the number of data pairs
04

Hypothesis Testing

Now, we have to perform a hypothesis test to determine if there is an increase in self-confidence scores after attending the course. The null hypothesis \(H_0:\mu_d = 0\) and alternative hypothesis \(H_1:\mu_d > 0\). Use the t statistic formula: \( t = \frac{\bar{d}-\mu_0}{s_d/\sqrt{n}} \). Compare the computed t-value with the critical value from the t-distribution table with the same degrees of freedom (n-1)

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Key Concepts

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

Confidence Interval
A confidence interval gives us a range of values that we believe contains the true mean of the population. In the context of hypothesis testing, specifically for paired differences, it helps us understand the range in which the true average change in scores likely falls.

A 95% confidence interval means that we are 95% confident that the actual mean of the paired differences is within this range. It's important because it provides an estimate that is more reliable than a single statistic, which might be misleading if it is affected by random sample variation.

When constructing a confidence interval for paired differences, we calculate the mean of the differences and their standard deviation. With these statistics, and using the standard z-score for a 95% confidence level (typically 1.96), we can compute the interval using the formula: \[ CI = \bar{d} \pm Z*\frac{s_d}{\sqrt{n}} \] where \( \bar{d} \) is the mean of the differences, \( s_d \) is the standard deviation of the differences, \( Z \) is the z-score, and \( n \) is the number of pairs.

This approach gives us a statistically backed boundary within which the true mean difference likely lies.
Paired Difference
In hypothesis testing, paired differences are useful when comparing two related sets of data. Here, 'paired' refers to the fact that the data points are connected, like the 'before' and 'after' scores of the same individual.

The paired difference for a single individual is the difference in their 'before' score and 'after' score. By focusing on these differences, we can control for the variability between individuals.

Suppose we have an employee with a 'before' score of 8 and an 'after' score of 10. The paired difference would be \[ \text{Paired Difference} = \text{Before Score} - \text{After Score} = 8 - 10 = -2 \].

This process is repeated for all participants to produce a set of paired differences. These differences tell us how much the scores typically changed, and they serve as the basis for calculating statistical metrics like the mean difference, which is crucial for subsequent confidence interval estimation and hypothesis testing.
Mean Score
The mean score in the context of paired differences refers to the average of all individual differences between 'before' and 'after' scores. It is a crucial indicator of the general effect that the intervention (in this case, the self-confidence course) had on the participants as a group.

To calculate the mean score of the paired differences (\( \bar{d}\)), you sum all the individual differences and divide by the number of participants:\[ \bar{d} = \frac{\sum{d}}{n}\]where \( \sum{d}\) is the total sum of all the differences, and \(n\) represents the number of paired observations.

The mean score provides a single value summary that reflects the typical amount of change in scores. If the mean is negative, it indicates a drop in scores; if it is positive, it indicates an improvement, which helps determine the impact of the course on boosting self-confidence.
Significance Level
The significance level in hypothesis testing is a threshold used to decide whether an observed effect is statistically meaningful. Commonly denoted by \(\alpha \), it represents the probability of rejecting the null hypothesis when it is actually true (Type I error).

In the exercise, a \(1\%\) significance level (or \(\alpha = 0.01\)) was used. This means we only consider results significant if the probability of observing the result, assuming the null hypothesis is true, is less than 1%.

Setting a low significance level like 1% indicates a stricter criterion for statistical significance, reducing the likelihood of a false positive result. To decide if the null hypothesis should be rejected, we compare the calculated p-value from the test against this significance level.
  • If the p-value is less than or equal to \(\alpha\), reject the null hypothesis.
  • If the p-value is greater than \(\alpha\), fail to reject the null hypothesis.
This process ensures that the increase in mean scores observed is unlikely due to random chance and more likely to be an actual effect of the self-confidence course.

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Most popular questions from this chapter

Briefly explain the meaning of independent and dependent samples. Give one example of each.

Manufacturers of two competing automobile models, Gofer and Diplomat, each claim to have the lowest mean fuel consumption. Let \(\mu_{1}\) be the mean fuel consumption in miles per gallon (mpg) for the Gofer and \(\mu_{2}\) the mean fuel consumption in mpg for the Diplomat. The two manufacturers have agreed to a test in which several cars of each model will be driven on a 100 -mile test run. Then the fuel consumption, in mpg, will be calculated for each test run. The average of the mpg for all 100 -mile test runs for each model gives the corresponding mean. Assume that for each model the gas mileages for the test runs are normally distributed with \(\sigma=2 \mathrm{mpg}\). Note that each car is driven for one and only one 100 -mile test run. a. How many cars (i.e., sample size) for each model are required to estimate \(\mu_{1}-\mu_{2}\) with a \(90 \%\) confidence level and with a margin of error of estimate of \(1.5 \mathrm{mpg}\) ? Use the same number of cars (i.e., sample size) for each model. b. If \(\mu_{1}\) is actually \(33 \mathrm{mpg}\) and \(\mu_{2}\) is actually \(30 \mathrm{mpg}\), what is the probability that five cars for each model would yield \(\bar{x}_{1} \geq \bar{x}_{2}\) ?

Maine Mountain Dairy claims that its 8-ounce low-fat yogurt cups contain, on average, fewer calories than the 8-ounce low-fat yogurt cups produced by a competitor. A consumer agency wanted to check this claim. A sample of 27 such yogurt cups produced by this company showed that they contained an average of 141 calories per cup. A sample of 25 such yogurt cups produced by its competitor showed that they contained an average of 144 calories per cup. Assume that the two populations are normally distributed with population standard deviations of \(5.5\) and \(6.4\) calories, repectively. a. Make a \(98 \%\) confidence interval for the difference between the mean number of calories in the 8-ounce low-fat yogurt cups produced by the two companies. b. Test at the \(1 \%\) significance level whether Maine Mountain Dairy's claim is true. c. Calculate the \(p\) -value for the test of part b. Based on this \(p\) -value, would you reject the null hypothesis if \(\alpha=.005 ?\) What if \(\alpha=.025\) ?

Gamma Corporation is considering the installation of governors on cars driven by its sales staff. These devices would limit the car speeds to a preset level, which is expected to improve fuel economy. The company is planning to test several cars for fuel consumption without governors for 1 week. Then governors would be installed in the same cars, and fuel consumption will be monitored for another week. Gamma Corporation wants to estimate the mean difference in fuel consumption with a margin of error of estimate of 2 mpg with a \(90 \%\) confidence level. Assume that the differences in fuel consumption are normally distributed and that previous studies suggest that an estimate of \(s_{d}=3 \mathrm{mpg}\) is reasonable. How many cars should be tested? (Note that the critical value of \(t\) will depend on \(n\), so it will be necessary to use trial and error.)

Two competing airlines, Alpha and Beta, fly a route between Des Moines, Iowa, and Wichita, Kansas. Each airline claims to have a lower percentage of flights that arrive late. Let \(p_{1}\) be the proportion of Alpha's flights that arrive late and \(p_{2}\) the proportion of Beta's tlights that arrive late. a. You are asked to observe a random sample of arrivals for each airline to estimate \(p_{1}-p_{2}\) with a \(90 \%\) confidence level and a margin of error of estimate of \(.05 .\) How many arrivals for each airline would you have to observe? (Assume that you will observe the same number of arrivals, \(n\), for each airline. To be sure of taking a large enough sample, use \(p_{1}=p_{2}=.50\) in your calculations for \(n .\) ) b. Suppose that \(p_{1}\) is actually \(.30\) and \(p_{2}\) is actually \(.23 .\) What is the probability that a sample of 100 flights for each airline ( 200 in all) would yield \(\hat{p}_{1} \geq \hat{p}_{2}\) ?
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