Final answer:
The 95% confidence interval for the mean amount spent on a child's birthday gift, with a sample mean of $53, standard deviation of $6, and sample size of 10, is approximately ($49.29, $56.71).
Step-by-step explanation:
To construct a 95% confidence interval for the mean amount spent on a child's birthday gift when the sample mean is $53, the standard deviation is $6, and the sample size is 10, one would typically use the t-distribution since the sample size is small and the population standard deviation is unknown. However, the question states that results appeared bell-shaped, thus if we presume the population is normally distributed, we could use the z-distribution. A 95% confidence interval can be found using the formula ± (z-score) * (standard deviation/sqrt(sample size)).
The z-score that corresponds to a 95% confidence level is approximately 1.96. We can then calculate the error margin as 1.96 * (6/sqrt(10)), which gives us an error margin of approximately 3.71. Adding and subtracting this error margin from the sample mean gives us the confidence interval: ($53 - 3.71, $53 + 3.71), or ($49.29, $56.71).