Question 1195939: Dear tutor please help me with stat’s questions, A film-processing company claims that the owners of digital cameras store on average more than fifteen pictures on their cameras, before the eventually download these to a computer or print. A random sample of 10 digital camera owners produced the data below on the number of pictures stored on their digital cameras.
25;6;22;26;31;18;13;20;14;2
a) Test this claim at the 10% level of significance.
b) Estimate with 95% confidence the mean number of pictures stored on digital cameras.
Answer by ElectricPavlov(122)   (Show Source):
You can put this solution on YOUR website!
**a) Hypothesis Testing**
1. **Define Hypotheses:**
* **Null Hypothesis (H0):** μ ≤ 15 (Mean number of pictures stored is less than or equal to 15)
* **Alternative Hypothesis (H1):** μ > 15 (Mean number of pictures stored is greater than 15)
2. **Calculate Sample Mean and Standard Deviation:**
* **Sample Mean (x̄):**
* x̄ = (25 + 6 + 22 + 26 + 31 + 18 + 13 + 20 + 14 + 2) / 10 = 17.1
* **Sample Standard Deviation (s):**
* Calculate using the formula: s = √[Σ(x - x̄)² / (n - 1)]
* s ≈ 8.29
3. **Calculate Test Statistic:**
* **t-statistic:**
* t = (x̄ - μ0) / (s / √n)
* where:
* x̄: Sample mean (17.1)
* μ0: Hypothesized population mean (15)
* s: Sample standard deviation (8.29)
* n: Sample size (10)
* t = (17.1 - 15) / (8.29 / √10)
* t ≈ 0.766
4. **Determine Critical Value:**
* **Degrees of Freedom (df):** n - 1 = 10 - 1 = 9
* **Significance Level (α):** 0.10
* **One-tailed t-distribution:**
* Find the critical t-value (t_critical) from a t-distribution table or using statistical software.
* For α = 0.10, df = 9, t_critical ≈ 1.383
5. **Decision Rule:**
* If the calculated t-statistic is greater than the critical t-value (t > t_critical), reject the null hypothesis.
* If the calculated t-statistic is less than or equal to the critical t-value (t ≤ t_critical), fail to reject the null hypothesis.
6. **Make a Decision:**
* Since 0.766 < 1.383, we fail to reject the null hypothesis.
7. **Conclusion:**
* There is not enough evidence at the 10% level of significance to support the claim that the mean number of pictures stored on digital cameras is greater than 15.
**b) Estimate the Mean Number of Pictures with 95% Confidence**
1. **Find Critical t-value:**
* For a 95% confidence level and df = 9, the critical t-value (t_critical) is approximately 2.262.
2. **Calculate Margin of Error:**
* Margin of Error = t_critical * (s / √n)
* Margin of Error = 2.262 * (8.29 / √10)
* Margin of Error ≈ 5.93
3. **Calculate Confidence Interval:**
* Lower Limit: x̄ - Margin of Error = 17.1 - 5.93 = 11.17
* Upper Limit: x̄ + Margin of Error = 17.1 + 5.93 = 23.03
4. **Conclusion:**
* We are 95% confident that the true mean number of pictures stored on digital cameras lies between 11.17 and 23.03.
**Note:**
* This analysis assumes that the number of pictures stored on digital cameras follows a normal distribution.
* The t-distribution is used because the population standard deviation is unknown and we are using the sample standard deviation.
**a) Hypothesis Testing**
1. **Hypotheses:**
* **Null Hypothesis (H₀):** μ ≤ 15 (The mean number of pictures stored is less than or equal to 15)
* **Alternative Hypothesis (H₁):** μ > 15 (The mean number of pictures stored is greater than 15)
2. **Calculate Sample Statistics:**
* **Sample Mean (x̄):**
* (25 + 6 + 22 + 26 + 31 + 18 + 13 + 20 + 14 + 2) / 10 = 17.1
* **Sample Standard Deviation (s):**
* Calculate using the formula:
* s = √[Σ(x - x̄)² / (n - 1)]
* s ≈ 8.21
3. **Calculate Test Statistic:**
* **t-statistic:**
* t = (x̄ - μ₀) / (s / √n)
* where:
* x̄ is the sample mean (17.1)
* μ₀ is the hypothesized population mean (15)
* s is the sample standard deviation (8.21)
* n is the sample size (10)
* t = (17.1 - 15) / (8.21 / √10)
* t ≈ 0.77
4. **Determine Critical Value:**
* **Degrees of Freedom:** df = n - 1 = 10 - 1 = 9
* **Significance Level:** α = 0.10 (one-tailed test)
* **Find the critical t-value** from a t-distribution table.
* For α = 0.10 and df = 9, the critical t-value is approximately 1.383.
5. **Decision Rule:**
* If the calculated t-statistic is greater than the critical t-value, reject the null hypothesis.
6. **Conclusion:**
* Since the calculated t-statistic (0.77) is less than the critical t-value (1.383), we **fail to reject the null hypothesis**.
* There is not enough evidence at the 10% significance level to support the claim that the mean number of pictures stored on digital cameras is greater than 15.
**b) Estimate the Mean with 95% Confidence**
1. **Find Critical t-value:**
* For a 95% confidence level and df = 9, the critical t-value (from a t-distribution table) is approximately 2.262.
2. **Calculate Margin of Error:**
* Margin of Error = t * (s / √n)
* Margin of Error = 2.262 * (8.21 / √10)
* Margin of Error ≈ 5.87
3. **Calculate Confidence Interval:**
* Lower Limit: x̄ - Margin of Error = 17.1 - 5.87 = 11.23
* Upper Limit: x̄ + Margin of Error = 17.1 + 5.87 = 22.97
**Conclusion:**
* We are 95% confident that the true mean number of pictures stored on digital cameras lies between 11.23 and 22.97.
**Note:**
* This analysis assumes that the number of pictures stored follows a normal distribution.
* The t-distribution is used because the population standard deviation is unknown and we are using the sample standard deviation.
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