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ACT Scores A random survey of 1000 students nationwide showed a mean ACT score of 21.4 . Ohio was not used. A survey of 500 randomly selected Ohio scores showed a mean of 20.8 . If the population standard deviation is \(3,\) can we conclude that Ohio is below the national average? Use \(\alpha=0.05 .\)

Short Answer

Expert verified
Yes, Ohio's mean ACT score is significantly below the national average.

Step by step solution

01

Set Up Hypotheses

We want to test if the mean ACT score for Ohio is less than the national mean. Thus, we set up the hypotheses as follows:- Null Hypothesis \( H_0: \mu_{Ohio} = \mu_{national} \)- Alternative Hypothesis \( H_a: \mu_{Ohio} < \mu_{national} \).
02

Identify Known Values

The known values from the problem are as follows:- National mean \( \mu_{national} = 21.4 \)- Ohio mean \( \bar{x}_{Ohio} = 20.8 \)- Population standard deviation \( \sigma = 3 \)- Sample size for Ohio, \( n = 500 \)- Significance level \( \alpha = 0.05 \).
03

Calculate the Standard Error

To calculate the standard error (SE) of the mean for the Ohio sample, use the formula:\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{500}}. \]Calculating this gives:\[ SE \approx 0.134. \]
04

Compute the Test Statistic

The test statistic \( z \) can be found using the formula:\[ z = \frac{\bar{x}_{Ohio} - \mu_{national}}{SE}. \]Plugging in the values:\[ z = \frac{20.8 - 21.4}{0.134} \approx -4.48. \]
05

Determine the Critical Value

For a left-tailed test at \( \alpha = 0.05 \), the critical value \( z_{critical} \) from the standard normal distribution is approximately \(-1.645\).
06

Make a Decision

Compare the computed test statistic \( z = -4.48 \) with the critical value \( z_{critical} = -1.645 \). Since \(-4.48\) is less than \(-1.645\), we reject the null hypothesis.

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Key Concepts

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

Understanding ACT Scores
ACT scores play a crucial role in the college admissions process in the United States. The ACT is a standardized test that measures a student's readiness for college. It encompasses four main sections: English, Mathematics, Reading, and Science Reasoning, with an optional Writing test. Each of these sections is scored on a scale of 1 to 36, and a composite score is calculated as the average of the four main sections.
  • Standardized nature: All students take the same test under similar conditions.
  • Comparison across populations: Scores are used to compare the academic abilities of students nationally.
When illustrating differences in regional or state performances, as in the case of Ohio, comparing mean scores can provide insights into educational trends and needs. The nationwide average, like the mean score of 21.4 in our problem context, is often used as a benchmark for these comparisons.
The Role of Standard Error
In statistics, the standard error serves as a measure of how much a sample mean is expected to fluctuate from the true population mean due to random sampling.
The standard error is essential when making inferences about a population from a sample.
  • Formula: The standard error (SE) can be calculated using the formula \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
  • Purpose: It helps quantify the uncertainty associated with the sample mean.
In our case, with a population standard deviation of 3 and a sample size of 500, the calculated SE is approximately 0.134. This means that on average, the Ohio sample mean could deviate by about 0.134 points from the true population mean due to sampling variability.
Understanding Population Standard Deviation
The population standard deviation, often represented by the Greek letter \( \sigma \), is a measure of the spread or dispersion of a set of values in a population.
It provides insight into the variability of individual data points within the population mean.
  • Formula: It is calculated as \( \sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{N}} \), where \( x_i \) are the individual data points, \( \mu \) is the mean, and \( N \) is the population size.
  • Impact: A smaller \( \sigma \) indicates that the data points are close to the mean, while a larger \( \sigma \) indicates greater variability.
In our exercise, we have a known population standard deviation of 3, which is used to calculate the standard error. Given the standard deviation, understanding how tightly or loosely the data is clustered around the mean can provide insights into data reliability.
Significance Level in Hypothesis Testing
The significance level, denoted by \( \alpha \), is a threshold that determines the cut-off point for deciding whether to reject the null hypothesis in hypothesis testing.
It represents the probability of making a Type I error, which is rejecting a true null hypothesis.
  • Common Levels: Typical \( \alpha \) values are 0.05, 0.01, or 0.10, with 0.05 being the most commonly used in business and academic research.
  • Interpretation: An \( \alpha = 0.05 \) implies a 5% risk of concluding that a difference exists when there is none.
In our exercise, a significance level of 0.05 was chosen to test whether Ohio's mean ACT score is less than the national average. By using this cut-off, we confidently conclude that Ohio's mean is significantly different from the national mean if the z-score falls beyond the critical value, which it did in this case.

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Most popular questions from this chapter

For Exercises 9 through \(24,\) perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Winter Temperatures A random sample of daily high temperatures in January and February is listed. At \(\alpha=0.05,\) can it be concluded that there is a difference in variances in high temperature between the two months? $$ \begin{array}{l|ccccccccccc}{\text { Jan. }} & {31} & {31} & {38} & {24} & {24} & {42} & {22} & {43} & {35} & {42} \\ \hline \text { Feb. } & {31} & {29} & {24} & {30} & {28} & {24} & {27} & {34} & {27}\end{array} $$

For Exercises 7 through \(27,\) perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Victims of Violence A random survey of 80 women who were victims of violence found that 24 were attacked by relatives. A random survey of 50 men found that 6 were attacked by relatives. At \(\alpha=0.10,\) can it be shown that the percentage of women who were attacked by relatives is greater than the percentage of men who were attacked by relatives?

Home Prices According to the almanac, the average sales price of a single- family home in the metropolitan Dallas/Ft. Worth/Irving, Texas, area is dollar 215,200. The average home price in Orlando, Florida, is dollar 198,000. The mean of a random sample of 45 homes in the Texas metroplex was dollar 216,000 with a population standard deviation of dollar 30,000 . In the Orlando, Florida, area a sample of 40 homes had a mean price of dollar 203,000 with a population standard deviation of dollar 32,500 . At the 0.05 level of significance, can it be concluded that the mean price in Dallas exceeds the mean price in Orlando? Use the \(P\) -value method.

Working Breath Rate Two random samples of 32 individuals were selected. One sample participated in an activity which simulates hard work. The average breath rate of these individuals was 21 breaths per minute. The other sample did some normal walking. The mean breath rate of these individuals was 14. Find the 90 \(\%\) confidence interval of the difference in the breath rates if the population standard deviation was 4.2 for breath rate per minute.

For these exercises, perform each of these steps. Assume that all variables are normally or approximately normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified and assume the variances are unequal. Tax-Exempt Properties A tax collector wishes to see if the mean values of the tax-exempt properties are different for two cities. The values of the tax- exempt properties for the two random samples are shown. The data are given in millions of dollars. At \(\alpha=0.05,\) is there enough evidence to support the tax collector's claim that the means are different? $$ \begin{array}{cccc|cccc}{} & {\text { City } \mathbf{A}} & {} & {} & {} & {\text { City } \mathbf{B}} & {} & {} \\ \hline 113 & {22} & {14} & {8} & {82} & {11} & {5} & {15} \\ {25} & {23} & {23} & {30} & {295} & {50} & {12} & {9} \\ {44} & {11} & {19} & {7} & {12} & {68} & {81} & {2} \\ {31} & {19} & {5} & {2} & {} & {20} & {16} & {4} & {5}\end{array} $$

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