Question 1167902: 1. From past records it is known that the average life of a battery used in a digital clock is 305 days. The lives of the batteries are normally distributed. The battery was recently modifies to last longer. Samples of 20 modified batteries were tested. It was discovered that the mean life was 311 days, ant the sample standard deviation was 12 days. At the 0.05 level of significance, did the modification increase the mean life of the battery?
a. State the null and alternate hypotheses.
b. Show the decision graphically.
c. Compute t, and reach a decision.
d. Briefly summary your findings.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! 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.
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