Question

A random sample of 17 hotels in Boston had an average nightly room rate of $165.40 with a sample standard deviation of $21.70. The critical value for a 98% confidence interval around this sample mean is ________.

Answer 1

The critical value for a 98% confidence interval with a sample size of 17 hotels is found using the t-distribution with 16 degrees of freedom. Typically, this critical value is approximately 2.583.

Step-by-step explanation:

The student is asking about constructing a 98% confidence interval for the average nightly hotel room rate based on a sample. To find the critical value for the confidence interval, we use the t-distribution since the sample size is less than 30 and the population standard deviation is unknown. The degrees of freedom (df) for our t-distribution is equal to the sample size minus one, which is 16 (17 - 1).

For a 98% confidence level, we look up the critical t-value corresponding to a 1% tail (since 98% confidence leaves 2% in the tails, split equally to 1% in each tail) and 16 degrees of freedom. You can find this value using a t-distribution table or a statistical software package. Typically, the critical t-value for 16 df and a 98% confidence level is approximately 2.583.

We can summarize the needed steps for someone looking for the critical value:

1. Identify the confidence level and the sample size.
2. Calculate the degrees of freedom (df = sample size - 1).
3. Look up the critical t-value in a t-distribution table or use statistical software, finding the value that corresponds to the df and the specified confidence level.