Question 1167902
This problem calls for a **one-sample t-test** to determine if the modification increased the mean life of the battery. We use a t-test because the population standard deviation is unknown, and we are working with a sample standard deviation. The problem states that the lives of the batteries are normally distributed, which is a key assumption for the t-test.

**Given Information:**

  * Known population average life ($\\mu\_0$): 305 days
  * Sample size ($n$): 20 modified batteries
  * Sample mean ($\\bar{x}$): 311 days
  * Sample standard deviation ($s$): 12 days
  * Level of significance ($\\alpha$): 0.05

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**a. State the null and alternate hypotheses.**

  * **Null Hypothesis ($H\_0$):** The modification did not increase the mean life of the battery (i.e., the mean life is still 305 days or less).
    $H\_0: \\mu \\le 305$ days
  * **Alternate Hypothesis ($H\_1$):** The modification increased the mean life of the battery (i.e., the mean life is greater than 305 days).
    $H\_1: \\mu \> 305$ days

This is a **one-tailed (right-tailed) test**.

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**b. Show the decision graphically.**

To show the decision graphically, we need the critical t-value.

  * Degrees of Freedom ($df$) = $n - 1 = 20 - 1 = 19$
  * Significance Level ($\\alpha$) = 0.05
  * Type of Test: One-tailed (right-tailed)

Looking up the t-distribution table for $df=19$ and $\\alpha=0.05$ (one-tailed), the critical t-value is approximately **1.729**.

**Graphical Representation:**

```
      ^ Probability Density
      |
      |          /--
      |         /   \
      |        /     \
      |       /       \
      |______/_________\____________
      -3    -2    -1    0     1    2    3
                              ^   ^
                             1.729 (Critical Value)
      ___________________________|_________________
               Fail to Reject H0 | Reject H0 (alpha = 0.05)
```

  * The t-distribution curve is shown.
  * The critical value of $t = 1.729$ defines the boundary of the rejection region.
  * The shaded area to the right of $1.729$ represents the rejection region (where the top 5% of the distribution lies).
  * If the calculated t-value falls into this shaded region, we reject $H\_0$. Otherwise, we fail to reject $H\_0$.

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**c. Compute t, and reach a decision.**

**1. Compute the t-value:**

The formula for the t-statistic is:
$t = \\frac{\\bar{x} - \\mu\_0}{s / \\sqrt{n}}$

Substitute the given values:
$\\bar{x} = 311$ days
$\\mu\_0 = 305$ days
$s = 12$ days
$n = 20$

$t = \\frac{311 - 305}{12 / \\sqrt{20}}$
$t = \\frac{6}{12 / 4.4721}$
$t = \\frac{6}{2.6833}$
$t \\approx 2.236$

**2. Reach a decision:**

  * Calculated t-value: $2.236$
  * Critical t-value: $1.729$

Since the calculated t-value ($2.236$) is greater than the critical t-value ($1.729$), it falls into the rejection region.

**Decision:** Reject the null hypothesis ($H\_0$).

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**d. Briefly summarize your findings.**

At the 0.05 level of significance, the statistical analysis provides **sufficient evidence** to conclude that the modification did indeed increase the mean life of the battery. The sample of 20 modified batteries showed a mean life of 311 days, which is significantly higher than the past average of 305 days, leading to the rejection of the null hypothesis.