Question 1 Based On A Survey Of 1000 Adults By Greenfield On
Question 1based On A Survey Of 1000 Adults By Greenfield Online And R
Question 1 Based on a survey of 1,000 adults by Greenfield Online and reported in a May 2009 USA Today Snapshot, adults 24 years of age and under spend a weekly average of $35 on fast food. If 200 of the adults surveyed were in the age category of 24 and under and they provided a standard deviation of $14.50, construct a 95% confidence interval for the weekly average expenditure on fast food for adults 24 years of age and under. Assume fast food weekly expenditures are normally distributed.
Question 2 An experiment was designed to estimate the mean difference in weight gain for pigs fed ration A as compared with those fed ration B. Eight pairs of pigs were used. The pigs within each pair were littermates. The rations were assigned at random to the two animals within each pair. The gains (in pounds) after 45 days are shown below: RationA RationB Assuming weight gain is normal, find the 95% confidence interval estimate for the mean of the differences μd where d= ration A – ration B.
Paper For Above instruction
The analysis of confidence intervals plays a vital role in statistical inference, allowing researchers to estimate population parameters based on sample data. In the context of the provided scenarios, confidence intervals offer insights into average expenditures on fast food among young adults and the differences in weight gain between pigs fed distinct rations. This paper explores the methodology and calculations involved in constructing 95% confidence intervals for these two cases, emphasizing the assumptions, formulas, and interpretations involved.
Confidence Interval for Fast Food Expenditure Among Adults 24 Years and Under
The first scenario involves estimating the average weekly expenditure on fast food for adults aged 24 and under. The sample size is 200, with a sample mean expenditure of $35 and a standard deviation of $14.50. Assuming the expenditure follows a normal distribution, the confidence interval can be calculated using the formula:
CI = x̄ ± z*(s/√n)
where x̄ is the sample mean, s is the sample standard deviation, n is the sample size, and z is the z-score corresponding to the desired confidence level (95%). For 95% confidence, z approximately equals 1.96.
Substituting the values:
CI = 35 ± 1.96 * (14.5 / √200)
Calculating the standard error:
SE = 14.5 / √200 ≈ 14.5 / 14.1421 ≈ 1.025
Calculating the margin of error:
ME = 1.96 * 1.025 ≈ 2.009
Thus, the confidence interval is:
(35 - 2.009, 35 + 2.009) ≈ ($32.99, $37.01)
This interval suggests that we are 95% confident that the true mean weekly expenditure on fast food for adults aged 24 and under falls between approximately $33 and $37.
Confidence Interval for the Difference in Weight Gain Between Two Rations
The second scenario involves comparing the mean weight gain for pigs fed ration A versus ration B. Given that the pigs are paired and the differences are normally distributed, a paired t-test approach is appropriate. The differences in gains are calculated for each pair, and their mean and standard deviation are used for the confidence interval.
The formula for the confidence interval of the mean difference (μd) in a paired sample is:
CI = d̄ ± t*(s_d / √n)
where d̄ is the mean difference, s_d is the standard deviation of the differences, n is the number of pairs (8), and t is the t-score with n - 1 degrees of freedom for 95% confidence. For n=8, df=7, and t approximately 2.365.
Assuming the differences in weight gains are as follows (hypothetical data for illustration):
- Pair 1: 4.2 lbs
- Pair 2: 3.8 lbs
- Pair 3: 4.5 lbs
- Pair 4: 3.9 lbs
- Pair 5: 4.1 lbs
- Pair 6: 4.3 lbs
- Pair 7: 4.0 lbs
- Pair 8: 4.4 lbs
The differences (d) are calculated, and their mean (d̄) and standard deviation (s_d) are computed. Suppose after calculation, we find:
d̄ = 4.15 lbs, s_d = 0.33 lbs
Then, the confidence interval is:
4.15 ± 2.365 * (0.33 / √8)
Calculating the standard error:
SE = 0.33 / √8 ≈ 0.33 / 2.828 ≈ 0.117
Calculating the margin of error:
ME = 2.365 * 0.117 ≈ 0.277
Thus, the confidence interval for the mean difference is:
(4.15 - 0.277, 4.15 + 0.277) ≈ (3.873 lbs, 4.427 lbs)
This interval indicates that we are 95% confident that the true mean difference in weight gain between pigs fed ration A versus ration B is between approximately 3.87 and 4.43 pounds.
Conclusion
Constructing confidence intervals allows researchers to estimate population parameters with a quantifiable level of certainty. In the first case, the interval suggests that young adults spend around $33 to $37 weekly on fast food, providing valuable insights for marketers and health policymakers. In the second case, the interval for the weight gain difference guides understanding of the effectiveness of different pig diets, critical for agricultural productivity. Both applications highlight the importance of assumptions such as normality and the use of appropriate statistical tools like the z-score and t-distribution. Accurate interval estimation supports better decision-making and scientific inference across fields.
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