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400 students were randomly sampled from a large university, and 289 said they did not get enough sleep. Conduct a hypothesis test to check whether this represents a statistically significant difference from \(50 \%\), and use a significance level of 0.01 .

Short Answer

Expert verified
Reject null hypothesis; significant difference from 50% exists.

Step by step solution

01

Define Hypotheses

Start by defining the null and alternative hypotheses. Here, the null hypothesis (\(H_0\)) states that the proportion of students who did not get enough sleep is equal to 50%, or \(p = 0.5\). The alternative hypothesis (\(H_a\)) states that the proportion is not equal to 50%, or \(p eq 0.5\).
02

Determine Sample Statistics

Calculate the sample proportion. From the problem, 289 out of 400 students reported insufficient sleep. The sample proportion \(\hat{p}\) is calculated as follows: \(\hat{p} = \frac{289}{400} = 0.7225\).
03

Calculate Standard Error

The standard error (SE) of the sample proportion is calculated using the formula \(SE = \sqrt{\frac{p(1-p)}{n}}\), where \(p=0.5\) and \(n=400\). Thus, \(SE = \sqrt{\frac{0.5(1-0.5)}{400}} = 0.025\).
04

Compute Test Statistic

Using the z-test, compute the test statistic as \(z = \frac{\hat{p} - p}{SE}\), where \(\hat{p} = 0.7225\), \(p = 0.5\), and \(SE = 0.025\). Thus, \(z = \frac{0.7225 - 0.5}{0.025} = 8.9\).
05

Determine Critical Value and Decision Rule

For a two-tailed test at a significance level of 0.01, the critical z-values are \(\pm 2.576\). If the calculated z-value falls outside of this range, we reject the null hypothesis.
06

Make a Decision

The calculated z-value is 8.9, which is greater than 2.576. Therefore, the result is significant, and we reject the null hypothesis \(H_0\).
07

State Conclusion

There is strong evidence at the 0.01 significance level to conclude that the proportion of students who do not get enough sleep is significantly different from 50%.

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Key Concepts

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

Sample Proportion
The sample proportion is a valuable concept in statistics, especially in hypothesis testing. Essentially, the sample proportion is an estimate of the actual proportion of a population. In this exercise, out of 400 students, 289 reported not having enough sleep. The sample proportion, denoted by \( \hat{p} \), is calculated as the number of successes (in this case, students not getting enough sleep) divided by the total sample size.
  • Formula: \( \hat{p} = \frac{x}{n} \)
  • Where \( x \) is the number of successes, and \( n \) is the total number of observations.

Applying this, we have \( \hat{p} = \frac{289}{400} = 0.7225 \). This result means that approximately 72.25% of the sampled students reported insufficient sleep.

Understanding the sample proportion is crucial because it serves as the foundation for calculating other important statistics like standard error and test statistics in hypothesis testing.
Standard Error
The standard error (SE) is a measure of the variability or dispersion of a sample statistic, like the sample proportion, from the true population proportion. It gives us an idea of how much sampling variability there is likely to be.
  • It is an essential component in hypothesis testing as it helps in constructing confidence intervals and evaluating the significance of tests.

For sample proportions, the standard error is calculated using the formula:
\[ SE = \sqrt{\frac{p(1-p)}{n}} \]
Where:
  • \( p \) is the assumed population proportion under the null hypothesis.
  • \( n \) is the number of observations in the sample.

In this exercise, \( p = 0.5 \) (50%), and \( n = 400 \), hence:
\[ SE = \sqrt{\frac{0.5 \times (1 - 0.5)}{400}} = 0.025 \]
This result indicates the standard deviation of the sample proportion, assuming that the true proportion of the population is 0.5.
Z-Test
The Z-Test is a type of statistical test that is used to determine whether there is a significant difference between sample and population means or proportions. It is particularly useful when the sample size is large.
  • In hypothesis testing, a Z-test can compare the sample proportion against a known population proportion to see if the difference is significant.

The formula for the Z-statistic in the context of testing a sample proportion against a population proportion is:
\[ z = \frac{\hat{p} - p}{SE} \]
Where:
  • \( \hat{p} \) is the sample proportion.
  • \( p \) is the hypothesized population proportion.
  • \( SE \) is the standard error of the sample proportion.

For the exercise, with \( \hat{p} = 0.7225 \), \( p = 0.5 \), and a standard error of 0.025, the Z-statistic is calculated as:
\[ z = \frac{0.7225 - 0.5}{0.025} = 8.9 \]
This high value suggests that the sample proportion is significantly different from the hypothesized population proportion.
Null Hypothesis
The null hypothesis is a fundamental concept in hypothesis testing. It is a statement that there is no effect or no difference, and it serves as the starting assumption in statistical testing.
  • The null hypothesis is symbolized by \( H_0 \), and is often the hypothesis that a researcher or scientist aims to test or disprove.

In this exercise, the null hypothesis is that the proportion of students who do not get enough sleep is equal to 50%, i.e., \( p = 0.5 \).

When conducting a hypothesis test, we determine whether there is sufficient statistical evidence in the sample to reject the null hypothesis.
Decision-making involves comparing the calculated test statistic (like the Z-score) with critical values.
If the test statistic falls in the rejection region beyond the critical values, the null hypothesis is rejected in favor of the alternative hypothesis, indicating a statistically significant difference.

In this case, since the calculated Z-value (8.9) exceeded the critical value (around 2.576 for a 0.01 significance level), the null hypothesis was rejected. This means the sample provides strong evidence that the proportion of students not getting enough sleep is different from 50%.

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Most popular questions from this chapter

A hospital administrator hoping to improve wait times decides to estimate the average emergency room waiting time at her hospital. She collects a simple random sample of 64 patients and determines the time (in minutes) between when they checked in to the ER until they were first seen by a doctor. A \(95 \%\) confidence interval based on this sample is (128 minutes, 147 minutes), which is based on the normal model for the mean. Determine whether the following statements are true or false, and explain your reasoning. (a) We are \(95 \%\) confident that the average waiting time of these 64 emergency room patients is between 128 and 147 minutes. (b) We are \(95 \%\) confident that the average waiting time of all patients at this hospital's emergency room is between 128 and 147 minutes. (c) \(95 \%\) of random samples have a sample mean between 128 and 147 minutes. (d) A \(99 \%\) confidence interval would be narrower than the \(95 \%\) confidence interval since we need to be more sure of our estimate. (e) The margin of error is 9.5 and the sample mean is 137.5 . (f) In order to decrease the margin of error of a \(95 \%\) confidence interval to half of what it is now, we would need to double the sample size. (Hint: the margin of error for a mean scales in the same way with sample size as the margin of error for a proportion.)

In each part below, there is a value of interest and two scenarios (I and II). For each part, report if the value of interest is larger under scenario I, scenario II, or whether the value is equal under the scenarios. (a) The standard error of \(\hat{p}\) when (I) \(n=125\) or (II) \(n=500\). (b) The margin of error of a confidence interval when the confidence level is (I) \(90 \%\) or (II) \(80 \%\). (c) The p-value for a Z-statistic of 2.5 calculated based on a (I) sample with \(n=500\) or based on a (II) sample with \(n=1000\). (d) The probability of making a Type 2 Error when the alternative hypothesis is true and the significance level is (I) 0.05 or (II) 0.10 .

Write the null and alternative hypotheses in words and using symbols for each of the following situations. (a) Since 2008 , chain restaurants in California have been required to display calorie counts of each menu item. Prior to menus displaying calorie counts, the average calorie intake of diners at a restaurant was 1100 calories. After calorie counts started to be displayed on menus, a nutritionist collected data on the number of calories consumed at this restaurant from a random sample of diners. Do these data provide convincing evidence of a difference in the average calorie intake of a diners at this restaurant? (b) The state of Wisconsin would like to understand the fraction of its adult residents that consumed alcohol

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