Problem 3
Question
A medical researcher believes that women typically have lower serum cholesterol than men. To test this hypothesis, he took a sample of four hundred seventy-six men between the ages of nineteen and forty-four and found their mean serum cholesterol to be \(189.0 \mathrm{mg} / \mathrm{dl}\) with a sample standard deviation of \(34.2\). A group of five hundred ninety-two women in the same age range averaged \(177.2 \mathrm{mg} / \mathrm{dl}\) and had a sample standard deviation of \(33.3\). Is the lower average for the women statistically significant? Set \(\alpha=0.05\). Assume the variances are equal.
Step-by-Step Solution
Verified Answer
The specific conclusion whether the average cholesterol level in women is significantly lower than men's will depend on the calculated t-value and its comparison with the critical t-value.
1Step 1: Define the Hypotheses
The null hypothesis (\(H_0\)) is that the average serum cholesterol level in men and women is the same. The alternative hypothesis (\(H_a\)) is that the average serum cholesterol level in women is lower than in men: \(H_0: \mu_{men} = \mu_{women}\) and \(H_a: \mu_{women} < \mu_{men}\)
2Step 2: Calculate the Test Statistic
The formula for the test statistic in a two-sample t-test is: \(t = \frac{(\bar{x_1} - \bar{x_2}) - (\mu_{1} - \mu_{2})}{\sqrt{\frac{{s_1}^2}{n_1} + \frac{{s_2}^2}{n_2}}}\) Where \(\bar{x_1}\) and \(\bar{x_2}\) are the sample means, \(\mu_{1}\) and \(\mu_{2}\) are the population means (which are both 0 for this test), \(s_1\) and \(s_2\) are the sample standard deviations, and \(n_1\) and \(n_2\) are the sample sizes. Substituting the given values into the formula gives: \(t = \frac{(189.0 - 177.2) - (0)}{\sqrt{\frac{{34.2}^2}{476} + \frac{{33.3}^2}{592}}}\)
3Step 3: Find the Critical T-Value
The critical t-value is a point beyond which we reject the null hypothesis. It can be found in the t-tables or calculated using a statistical calculator or software. For an alpha level of 0.05 and degrees of freedom \(df = n_1 + n_2 - 2 = 476 + 592 - 2 = 1066\), the critical t-value (one-sided) is approximately 1.645.
4Step 4: Compare the Test Statistic to the Critical T-Value
If the calculated t-value is greater than the critical t-value, we reject the null hypothesis. After calculating the t-value, compare it with the critical t-value.
5Step 5: Conclude the Hypothesis Test
Based on the comparison, if the calculated t-value is found to be greater than the critical t-value, conclude that the difference in average serum cholesterol between men and women is statistically significant. If not, conclude that the difference is not significant.
Key Concepts
t-testnull hypothesissample standard deviationcritical t-value
t-test
A t-test is a statistical method that allows us to determine whether there is a significant difference between the means of two groups. It helps us test if any observed differences in sample means reflect real changes in the population or if they just occurred by random chance.
In the context of the exercise, a two-sample t-test is used because we are comparing the serum cholesterol levels between two different groups: men and women.
What makes the t-test unique is that it takes into account the size and standard deviations of both samples, giving us a more reliable result even when sample sizes are different. When performing a t-test, the calculated test statistic can inform us how far the sample mean is from the population mean, considering the sample size and variation.
This test helps us decide whether any observed difference is significant enough to reject the null hypothesis.
In the context of the exercise, a two-sample t-test is used because we are comparing the serum cholesterol levels between two different groups: men and women.
What makes the t-test unique is that it takes into account the size and standard deviations of both samples, giving us a more reliable result even when sample sizes are different. When performing a t-test, the calculated test statistic can inform us how far the sample mean is from the population mean, considering the sample size and variation.
- In a t-test, the formula involves differences in sample means subtracted by differences in population means.
- We divide by a calculated standard error to standardize the result.
This test helps us decide whether any observed difference is significant enough to reject the null hypothesis.
null hypothesis
The null hypothesis (\(H_0\)) is a fundamental concept in hypothesis testing. It assumes that any kind of difference or significance you observe in a data set is due to chance. In simpler terms, the null hypothesis suggests that there is no actual effect or difference.
In the medical research example, the null hypothesis states that there is no difference in the average serum cholesterol levels between men and women. Mathematically, it is represented as \(H_0: \mu_{men} = \mu_{women}\).
If our statistical test indicates that the null hypothesis is less likely to be true, we may reject it in favor of an alternative hypothesis (\(H_a\)). The alternative hypothesis for this exercise is that women have lower serum cholesterol than men (\(H_a: \mu_{women} < \mu_{men}\)).
Rejecting the null hypothesis suggests the data provide enough evidence of a significant difference.
In the medical research example, the null hypothesis states that there is no difference in the average serum cholesterol levels between men and women. Mathematically, it is represented as \(H_0: \mu_{men} = \mu_{women}\).
If our statistical test indicates that the null hypothesis is less likely to be true, we may reject it in favor of an alternative hypothesis (\(H_a\)). The alternative hypothesis for this exercise is that women have lower serum cholesterol than men (\(H_a: \mu_{women} < \mu_{men}\)).
- Testing the null hypothesis typically involves calculating a test statistic.
- We then compare this statistic to a critical value determined by the specified significance level (\(\alpha\)).
Rejecting the null hypothesis suggests the data provide enough evidence of a significant difference.
sample standard deviation
Sample standard deviation is an essential part of statistical calculations as it measures the amount of variation or dispersion present in a dataset. It indicates how much the values in the sample deviate from the mean on average.
In the exercise, the sample standard deviations are crucial for calculating the test statistic in the t-test. Each group's variation is considered to determine if the difference in cholesterol levels is significant.
The formula used to find the standard deviation of a sample is slightly different from that for a population, due to the "n-1" term in the denominator. This adjustment, known as Bessel's correction, makes the sample standard deviation an unbiased estimator of the population standard deviation:
Including these values in the test statistic calculation allows us to factor in both the amount of data and the consistency within each group's cholesterol levels.
In the exercise, the sample standard deviations are crucial for calculating the test statistic in the t-test. Each group's variation is considered to determine if the difference in cholesterol levels is significant.
The formula used to find the standard deviation of a sample is slightly different from that for a population, due to the "n-1" term in the denominator. This adjustment, known as Bessel's correction, makes the sample standard deviation an unbiased estimator of the population standard deviation:
- For men, the sample standard deviation is 34.2 mg/dl.
- For women, it is slightly lower at 33.3 mg/dl.
Including these values in the test statistic calculation allows us to factor in both the amount of data and the consistency within each group's cholesterol levels.
critical t-value
The critical t-value is a threshold used to compare against the calculated t-value in hypothesis testing. It serves as a cutoff point in deciding whether to reject the null hypothesis.
The critical t-value is determined based on the chosen significance level, \(\alpha\), the type of test (one-tailed or two-tailed), and the degrees of freedom (df). For the exercise, these values are \(\alpha = 0.05\) and degrees of freedom \(df = 1066\).
A critical t-value can be found in a t-distribution table or by using statistical software. For this specific one-tailed test with \(\alpha = 0.05\), the critical t-value is approximately 1.645.
Thus, the critical t-value plays a vital role in determining the statistical significance of the findings.
The critical t-value is determined based on the chosen significance level, \(\alpha\), the type of test (one-tailed or two-tailed), and the degrees of freedom (df). For the exercise, these values are \(\alpha = 0.05\) and degrees of freedom \(df = 1066\).
A critical t-value can be found in a t-distribution table or by using statistical software. For this specific one-tailed test with \(\alpha = 0.05\), the critical t-value is approximately 1.645.
- If the calculated t-value exceeds the critical t-value, we reject the null hypothesis.
- Conversely, if it falls short, we fail to reject the null hypothesis, indicating that any observed difference might not be significant.
Thus, the critical t-value plays a vital role in determining the statistical significance of the findings.
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