Math Problem Statement

A study of the effect of television commercials on 12-year-old children measured their attention span, in seconds. The commercials were for clothes, food, and toys.

Clothes Food Toys 28 38 53 16 41 44 51 42 48 35 52 56 28 47 63 31 42 53 17 34 48 31 43 58 20 57 47 47 51 44 51 54 Compute the 95% confidence intervals that estimate the difference between each pair of means. (Negative amount should be indicated by a minus sign. Round your answers to 2 decimal places.)

Solution

To compute the 95% confidence intervals for the differences between each pair of means, we need to follow these steps:

Step 1: Calculate the means for each category (Clothes, Food, and Toys).

We will compute the mean of the attention span for each group (Clothes, Food, and Toys).

Step 2: Compute the standard deviations for each group.

We need to calculate the standard deviation for the three groups to assess the variability in the data.

Step 3: Calculate the standard error for each pair of means.

The standard error of the difference between two means is given by the formula:

SE=(SD12n1)+(SD22n2)SE = \sqrt{\left(\frac{SD_1^2}{n_1}\right) + \left(\frac{SD_2^2}{n_2}\right)} Where:

  • SD1SD_1 and SD2SD_2 are the standard deviations of the two groups.
  • n1n_1 and n2n_2 are the sample sizes of the two groups.

Step 4: Compute the margin of error.

The margin of error is computed using the following formula:

ME=t×SEME = t \times SE

Where tt is the critical value from the t-distribution for a 95% confidence level, based on the degrees of freedom. For large samples (which we approximate when n30n \geq 30), the tt-value is close to 1.96.

Step 5: Compute the confidence interval for the difference between each pair of means.

The confidence interval is given by:

Confidence Interval=(Mean DifferenceME,Mean Difference+ME)\text{Confidence Interval} = (\text{Mean Difference} - \text{ME}, \text{Mean Difference} + \text{ME})

Now, let's compute these values.

I will start by calculating the mean, standard deviation, and confidence intervals for the data you provided.### Summary of Results:

  • Mean values:

    • Clothes: 30.40 seconds
    • Food: 45.75 seconds
    • Toys: 52.22 seconds
  • 95% Confidence Intervals for the differences between means:

    1. Clothes vs. Food: (-24.99, -5.71)
    2. Clothes vs. Toys: (-31.54, -12.11)
    3. Food vs. Toys: (-12.97, 0.03)

These confidence intervals represent the estimated difference between the means of attention span for the given commercials. A negative value indicates that the first group (e.g., Clothes) has a lower mean than the second group (e.g., Food).

Would you like more details or have any questions?

Related Questions:

  1. How is the t-critical value for confidence intervals determined?
  2. What are the effects of standard deviation on the width of confidence intervals?
  3. How can you interpret a confidence interval that includes zero?
  4. Why are the confidence intervals for "Clothes vs. Toys" wider than "Food vs. Toys"?
  5. How does sample size affect the precision of a confidence interval?

Tip:

Confidence intervals become narrower with larger sample sizes or lower variability, improving the precision of your estimates.

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Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Intervals
Means
Standard Deviation
Standard Error

Formulas

Mean = Sum of values / Number of values
Standard Deviation = sqrt(Sum of (xi - mean)^2 / (n - 1))
Standard Error (SE) = sqrt((SD1^2 / n1) + (SD2^2 / n2))
Confidence Interval = Mean Difference ± (t * SE)

Theorems

Central Limit Theorem
T-distribution

Suitable Grade Level

Grades 11-12 (Advanced Statistics)