Question 1195939

**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.