/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 A test of the breaking strengths... [FREE SOLUTION] | 91影视

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A test of the breaking strengths of two different types of cables was conducted using samples of \(n_{1}=n_{2}=100\) pieces of each type of cable\begin{tabular}{ll} \hline Cable I & Cable II \\ \hline \(\bar{x}_{1}=1925\) & \(\bar{x}_{2}=1905\) \\ \(s_{1}=40\) & \(s_{2}=30\) \end{tabular} Do the data provide sufficient evidence to indicate a difference between the mean breaking strengths of the two cables? Use \(\alpha=.05 .\)

Short Answer

Expert verified
Answer: Yes, there is sufficient evidence to indicate a difference between the mean breaking strengths of the two cables at a 0.05 significance level.

Step by step solution

01

State the hypotheses

We are trying to determine if there is a difference between the mean breaking strengths of the two cables. This means that our null and alternative hypotheses will be as follows: Null Hypothesis (H鈧): 碌鈧 = 碌鈧 (There is no difference between the mean breaking strengths of the two cables) Alternative Hypothesis (H鈧): 碌鈧 鈮 碌鈧 (There is a difference between the mean breaking strengths of the two cables)
02

Determine the test statistic

To perform the hypothesis test, we will use the two-sample t-test. The test statistic formula is: $$ t = \frac{(\bar{x}_1 - \bar{x}_2) - 0} {s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} $$ Where: - \(s_p\) represents the pooled standard deviation - \(s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}\) We are given the necessary data: - \(\bar{x}_1 = 1925\) - \(\bar{x}_2 = 1905\) - \(n_1 = n_2 = 100\) - \(s_1 = 40\) - \(s_2 = 30\) First, we calculate the pooled standard deviation: $$ s_p = \sqrt{\frac{(100 - 1)(40)^2 + (100 - 1)(30)^2}{100 + 100 - 2}} $$ $$ s_p = \sqrt{\frac{3952000}{198}} $$ $$ s_p \approx 35.59 $$ Next, we can find the test statistic (t): $$ t = \frac{(1925 - 1905) - 0} {35.59 \sqrt{\frac{1}{100} + \frac{1}{100}}} $$ $$ t \approx \frac{20}{5.04} $$ $$ t \approx 3.97 $$
03

Find the critical value and determine the p-value

Now, we will find the critical value for our t-distribution given our significance level (伪 = 0.05) and degrees of freedom: Degrees of Freedom (df) = \(n_1 + n_2 - 2 = 100 + 100 - 2 = 198\) Using a t-table or software, we can find the critical value for a two-tailed t-test (because our alternative hypothesis is 碌鈧 鈮 碌鈧): Critical Value (t) 鈮 1.972 To find p-value, we will use the t-distribution: Since our test statistic is 3.97, which is greater than the critical value of 1.972, our p-value will be less than 伪 = 0.05.
04

Decide to reject or fail to reject the null hypothesis

Since our p-value is less than 伪 = 0.05, we reject the null hypothesis (H鈧: 碌鈧 = 碌鈧).
05

Conclusion

There is sufficient evidence to indicate a difference between the mean breaking strengths of the two cables, at a 0.05 significance level.

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Key Concepts

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

Two-Sample t-Test
The two-sample t-test is a widely used statistical method to determine if there are significant differences between the means of two independent groups. In the context of the exercise, we are evaluating whether two different types of cables have different mean breaking strengths. To apply this test, we require two sample means, their standard deviations, and sample sizes.

Here's how the two-sample t-test works in practice: Assume you have data from two samples. First, calculate the means and standard deviations of each sample, then use these to compute the test statistic. The test statistic essentially measures the distance, in standard error units, between the two sample means. If the test statistic falls beyond a certain critical value in the t-distribution, it suggests that the observed difference between sample means is not likely due to random chance, hence the means of the two populations are significantly different.

It's crucial to check assumptions such as the independence of samples and the normal distribution of data within both groups when using this test. If these assumptions are not met, the results of the t-test may not be reliable.
Null and Alternative Hypotheses
The null and alternative hypotheses are the backbone of hypothesis testing, including the two-sample t-test. To understand this foundation, let's define both terms:
  • The null hypothesis (\(H_0\)) posits that there is no effect or no difference between groups. In our exercise, it assumes the mean breaking strengths of the two cables are equal (\( \bar{x}_1 = \bar{x}_2 \(H_0\).
  • The alternative hypothesis (\(H_a\(H_a\).

    When performing a hypothesis test, we are trying to obtain evidence to support rejecting the null hypothesis. Essentially, we're seeking to prove the alternative hypothesis by demonstrating that any observed difference is statistically significant and not merely due to random chance within the samples tested. If the results indicate that the null hypothesis is unlikely (typically with a p-value less than the chosen significance level, such as 0.05), we reject it in favor of the alternative hypothesis.
Pooled Standard Deviation
The pooled standard deviation is a measure used in statistics to estimate the combined standard deviation of two independent samples, assuming that they have the same variance. It is used in the two-sample t-test when we are comparing the means of these samples.

To calculate the pooled standard deviation, you follow a specific formula:\[\begin{equation}s_p = \[s_p = \$; where \( s_{1}\) and \(s_{2}\) are the standard deviations of the first and second samples, respectively, and \(n_{1}\) and \(n_{2}\) are the sample sizes. This pooled figure represents an average spread of all data points around their group means, weighted by their degrees of freedom.

By using the pooled standard deviation in our test statistic calculation, we incorporate an estimate of the common variance between the samples. This helps to strengthen our two-sample t-test analysis. In the exercise, we calculate the pooled standard deviation using the given sample sizes and standard deviations of the breaking strengths, which then feeds into the t-statistic formula used to ultimately assess the difference between cable breaking strengths.

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Most popular questions from this chapter

An experimenter has prepared a drug-dose level that he claims will induce sleep for at least \(80 \%\) of people suffering from insomnia. After examining the dosage we feel that his claims regarding the effectiveness of his dosage are too high. In an attempt to disprove his claim, we administer his prescribed dosage to 50 insomniacs and observe that 37 of them have had sleep induced by the drug dose. Is there enough evidence to refute his claim at the \(5 \%\) level of significance?

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