Please help!! I have no idea how to do this question.
14. A sample of 40 CD’s from your collection showed a mean length of 52.74 minutes with a standard deviation of 13.21 minutes
a. Construct the 95 percent confidence interval for the mean.
b. Construct the 95 percent confidence interval for the mean using the t-value rather than the z-value. Comment on why the interval is larger when you use the t-value.
c. Why might normality of the data set be in question here? Comment on your answer.
To solve this question, we need to calculate the confidence interval for the mean using both the z-value and the t-value. We also need to comment on why the t-value interval is larger. Additionally, we need to discuss why normality of the data set might be in question. Let's break it down step by step:
a. To construct the 95% confidence interval for the mean using the z-value, we can use the following formula:
Confidence Interval = Sample Mean ± (Z-value * Standard Deviation / sqrt(n))
In this case, the sample mean is 52.74 minutes, the standard deviation is 13.21 minutes, and the sample size is 40. The Z-value for a 95% confidence interval is approximately 1.96 (consulting the standard normal Z-table).
Plugging in the values, we get:
Confidence Interval = 52.74 ± (1.96 * 13.21 / sqrt(40))
Solving this equation will give you the confidence interval.
b. To construct the 95% confidence interval for the mean using the t-value, we first need to find the degrees of freedom. Since the sample size is 40, the degrees of freedom is 40 - 1 = 39.
Next, we need to find the appropriate t-value for a 95% confidence interval and 39 degrees of freedom. You can use a t-table or statistical software to find this value. It is usually larger than the z-value used in part a.
Once you have the t-value, you can use the same formula as in part a, but with the t-value substituted for the z-value:
Confidence Interval = Sample Mean ± (T-value * Standard Deviation / sqrt(n))
Solving this equation will give you the confidence interval.
The reason the t-value interval is larger than the z-value interval is that the t-distribution has fatter tails compared to the normal distribution. This means that the confidence interval based on the t-distribution will be wider, allowing for more variability in the data.
c. Normality of the data set might be in question if the data does not follow a normal distribution. Several factors can affect the normality of the data set, such as outliers, skewness, or small sample size. In this case, since we are considering the lengths of CDs, it is possible that the distribution of the lengths is not perfectly normal. It could be skewed if there is a significant number of CDs with unusual lengths, or it could be multimodal if there are distinct groups of CDs with different lengths. Before making any assumptions about normality, it would be helpful to plot a histogram or perform a normality test on the data.