In previous years the average number of patients per hour at...

In previous years the average number of patients per hour at a hospital emergency room on weekends exceeded the average on weekdays by 6.3 visits per hour. A hospital administrator believes that the current weekend mean exceeds the weekday mean by fewer than 6.3 hours. a. Construct the $99 \%$ confidence interval for the difference in the population means based on the following data, derived from a study in which 30 weekend and 30 weekday one-hour periods were randomly selected and the number of new patients in each recorded. b. Test at the $5 \%$ level of significance whether the current weekend mean exceeds the weekday mean by fewer than 6.3 patients per hour. c. Compute the observed significance of the test.

In previous years the average number of patients per hour at a hospital emergency room on weekends
exceeded the average on weekdays by 6.3 visits per hour. A hospital administrator believes that the current
weekend mean exceeds the weekday mean by fewer than 6.3 hours.
a. Construct the $99 \%$ confidence interval for the difference in the population means based on the
following data, derived from a study in which 30 weekend and 30 weekday one-hour periods
were randomly selected and the number of new patients in each recorded.
b. Test at the $5 \%$ level of significance whether the current weekend mean exceeds the weekday
mean by fewer than 6.3 patients per hour.
c. Compute the observed significance of the test.

Key Concepts

Confidence Interval for Difference in Means

A confidence interval for the difference in means is a range computed from sample data that is likely to contain the true difference between two population means. It is constructed using the sample means, sample standard deviations, and the relevant critical value from the t-distribution (or z-distribution, under large samples) corresponding to a predetermined confidence level. This interval provides an estimation of the parameter with an associated level of confidence, meaning that if the study were repeated multiple times, a certain percentage (e.g., 99% for a 99% confidence interval) of the intervals constructed would contain the true difference.

Two-Sample t-Test

The two-sample t-test is a statistical method used to determine whether there is a statistically significant difference between the means of two independent groups. This test is particularly useful when the sample sizes are small and the population standard deviations are unknown. It works by comparing the observed difference between sample means to the expected difference under the null hypothesis, and uses the t-distribution to account for variability and sample size.

Hypothesis Testing

Hypothesis testing is a formal procedure used in statistics to make inferences about population parameters based on sample data. It involves setting up a null hypothesis, which represents a baseline or no-effect scenario, and an alternative hypothesis, which reflects the effect or difference being tested. The test statistic is calculated from the data and compared against a critical value from a relevant probability distribution to determine whether there is enough evidence to reject the null hypothesis in favor of the alternative.

Significance Level

The significance level, often denoted as alpha (?), is the threshold used in hypothesis testing to determine when to reject the null hypothesis. It represents the probability of incorrectly rejecting the null hypothesis (a Type I error). Common choices for the significance level include 0.05 (5%), 0.01 (1%), etc. The chosen ? level influences the critical value against which the test statistic is compared.

p-Value

The p-value is the probability of obtaining a test statistic as extreme as, or more extreme than, the observed value, assuming that the null hypothesis is true. It is used to assess the strength of the evidence against the null hypothesis. A smaller p-value indicates stronger evidence against the null hypothesis, and if the p-value is less than the pre-determined significance level, the null hypothesis is rejected.

Transcript

00:01 The following is a solution to number 15, and a hospital administrator wants to look at the average number of patients per hour at a hospital emergency room on the weekends, and we're going to compare it to the same thing for weekdays, and we want to see if it's exceeding 6 .3 visits per hour, which is kind of crazy, but i guess i understand.

00:18 The weekend, so they selected 30 weekend days, and i found the mean of 13 .8 patients per hour, and then the standard deviation 3 .1 weekday is 30 with the, x bar or sample mean is 8 .6 and a patience per hour and then the standard deviation 2 .7.

00:37 And so we're asked to find a 99 % confidence interval of that to see that, you know, the difference of the two.

00:44 And so i'm going to use the ti84 here.

00:46 If we go to stat and then go to test, we're going to go to that ninth option there.

00:52 And we need to make sure that we, you know, punch in these data values.

00:55 So s1 is in 3 .1, s2 was 2 .2.

01:00 Now i know it says sigma, but it's this because the sample size is large enough, we can just use the sample standard deviations and this mean is 13 .8 and then this first sample size was 30 and then the second mean was 8 .6 and then that sample size was 30 and we're asked to find the 99 % so 0 .99 and we calculate and that gives us 3 .267 to 7 .1333.

01:29 Okay, so let's write that down so we have 3 .2667 all the way up to 7 .13333.

01:40 That's the 99 % confidence interval of the difference between weekdays and week, sorry, weekends and week days.

01:48 Now we're going to test at the 5 % level of confidence if the difference is 6 .3 or less than 6 .3.

01:55 Now i can't use a calculator on this one, unfortunately, because the calculator only does it if it's, zero.

02:03 So the difference is zero and this is the difference of 6 .3.

02:06 So kind of unfortunate, but that's okay.

02:08 We can use the formula.

02:09 So it's the difference of the sample means, which is 13 .8 minus 8 .6 minus the hypothesized difference, which is 16, i'm sorry, 6 .3.

02:19 And then we divide that by the standard deviation or standard error, which is the standard deviation squared over the sample size plus standard deviation squared over that sample size.

02:31 And whenever you punch this into the calculator very carefully, you should get negative 1 .466 as your test statistic.

02:38 So what do we do with that? we find the p value, and the p value is the probability that it is less than, since the alternate hypothesis is less than, less than 1 .466.

02:50 So if we go to second distribution and we're going to go to the normal cdf, the lower bound is going to be negative some large number, at least in the negative sense...

Please give Ace some feedback