An engineer wanted to know whether the strength of two different concrete mix designs differed significantly. He randomly selected 9 cylinders, measuring 6 inches in diameter and 12 inches in height, into which mixture 67-0-301 was poured. After 28 days, he measured the strength (in pounds per square inch) of the cylinder. He also randomly selected 10 cylinders of mixture 67-0-400 and performed the same test. The results are as follows:
Mixture 67-0-301 {3,960 4,090 3,100 3,830 3,200 3,780 4,080 4,040 2,940} Mixture 67-0-400
{4,070 4,890 5,020 4,330 4,640 5,220 4,190 3,730 4,120 4,620}
       determine whether mixture 67-0-400 is stronger than mixture 67-
0-301 at the a=0.05 level of significance?

To determine whether the strength of mixture 67-0-400 is significantly stronger than mixture 67-0-301, we can perform a **two-sample t-test**. This will help us assess whether the mean strength of the two concrete mixtures differs significantly.

Here are the steps to follow:

### 1. **State the hypotheses:**

- **Null Hypothesis (\(H_0\))**: The mean strength of the two mixtures is the same.
  \[
  H_0: \mu_{\text{301}} = \mu_{\text{400}}
  \]
- **Alternative Hypothesis (\(H_A\))**: The mean strength of mixture 67-0-400 is greater than the mean strength of mixture 67-0-301.
  \[
  H_A: \mu_{\text{400}} > \mu_{\text{301}}
  \]

This is a **one-tailed t-test**, since we're only interested in whether mixture 67-0-400 is stronger (i.e., has a higher mean strength).

### 2. **Calculate the means and standard deviations:**

- Let’s calculate the sample means (\(\bar{x}_{301}\) and \(\bar{x}_{400}\)) and sample standard deviations (\(s_{301}\) and \(s_{400}\)) for each group.

### 3. **Compute the test statistic:**

The formula for the t-statistic for two independent samples is:

\[
t = \frac{\bar{x}_{400} - \bar{x}_{301}}{\sqrt{\frac{s_{301}^2}{n_{301}} + \frac{s_{400}^2}{n_{400}}}}
\]

Where:
- \(\bar{x}_{301}\) and \(\bar{x}_{400}\) are the sample means,
- \(s_{301}\) and \(s_{400}\) are the sample standard deviations,
- \(n_{301}\) and \(n_{400}\) are the sample sizes (9 and 10, respectively).

### 4. **Determine the degrees of freedom:**

We use the following formula for the degrees of freedom:

\[
df = \frac{\left(\frac{s_{301}^2}{n_{301}} + \frac{s_{400}^2}{n_{400}}\right)^2}{\frac{\left(\frac{s_{301}^2}{n_{301}}\right)^2}{n_{301}-1} + \frac{\left(\frac{s_{400}^2}{n_{400}}\right)^2}{n_{400}-1}}
}
\]

### 5. **Compare the calculated t-value with the critical t-value:**

At a significance level of \(\alpha = 0.05\), we will compare the computed t-statistic with the critical t-value from the t-distribution table for the calculated degrees of freedom.

### 6. **Make a decision:**

If the t-statistic exceeds the critical t-value, we reject the null hypothesis and conclude that mixture 67-0-400 is significantly stronger than mixture 67-0-301. If not, we fail to reject the null hypothesis.

Let me calculate these values for you.Here are the calculated results:

- **Mean of mixture 67-0-301**: 3668.89 psi
- **Mean of mixture 67-0-400**: 4483.00 psi
- **Standard deviation of mixture 67-0-301**: 458.52 psi
- **Standard deviation of mixture 67-0-400**: 473.66 psi

The test statistic is:
- **t-statistic**: 3.804

The critical t-value for \(\alpha = 0.05\) and the calculated degrees of freedom is:
- **t-critical**: 1.740

The p-value associated with this t-statistic is:
- **p-value**: 0.00072

### Conclusion:
Since the t-statistic (3.804) is greater than the critical t-value (1.740) and the p-value (0.00072) is less than 0.05, we **reject the null hypothesis**. This means that mixture 67-0-400 is significantly stronger than mixture 67-0-301 at the 0.05 level of significance.

Would you like more details on the calculations or explanation? 

Here are 5 related questions:
1. What assumptions are made in a two-sample t-test?
2. How do you interpret a p-value in hypothesis testing?
3. What is the difference between a one-tailed and two-tailed t-test?
4. How can we check if the variances are equal between two groups?
5. What is the effect of sample size on the results of a t-test?

**Tip**: Always ensure that the assumptions of normality and equal variances (if using a standard t-test) are met before applying the test!

An engineer wanted to know whether the strength of two different concrete mix designs differed significantly. He randomly selected 9 cylinders, measuring 6 inches in diameter and 12 inches in height, into which mixture 67-0-301 was poured. After 28 days, he measured the strength (in pounds per square inch) of the cylinder. He also randomly selected 10 cylinders of mixture 67-0-400 and performed the same test. The results are as follows: Mixture 67-0-301 {3,960 4,090 3,100 3,830 3,200 3,780 4,080 4,040 2,940} Mixture 67-0-400 {4,070 4,890 5,020 4,330 4,640 5,220 4,190 3,730 4,120 4,620} Determine whether mixture 67-0-400 is stronger than mixture 67-0-301 at the a=0.05 level of significance?

Perform a two-sample t-test to compare the strength of concrete mixtures 67-0-301 and 67-0-400. Using sample data, we test whether the mean strength of mixture 67-0-400 is significantly higher than that of mixture 67-0-301 at a 0.05 significance level. This analysis includes key statistical formulas and hypothesis testing concepts.

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