Nursing Magazine reported results of a survey of more than...

Nursing Magazine reported results of a survey of more than 1800 nurses across the country concerning job satisfaction and retention. Nurses from magnet hospitals (hospitals that successfully attract and retain nurses) describe the staffing situation in their units as follows: A survey of 500 nurses from nonmagnet hospitals gave the following responses to the staffing situation. Do the data indicate that the nurses from the nonmagnet hospitals have a different distribution of opinions? Use $\alpha=0.05$ a. Solve using the $p$ -value approach. b. Solve using the classical approach.

Nursing Magazine reported results of a survey of more than 1800 nurses across the country concerning job satisfaction and retention. Nurses from magnet hospitals (hospitals that successfully attract and retain nurses) describe the staffing situation in their units as follows:
A survey of 500 nurses from nonmagnet hospitals gave the following responses to the staffing situation.
Do the data indicate that the nurses from the nonmagnet hospitals have a different distribution of opinions? Use $\alpha=0.05$
a. Solve using the $p$ -value approach.
b. Solve using the classical approach.

Key Concepts

Hypothesis Testing

Hypothesis testing is a systematic method used to determine whether there is enough statistical evidence in a sample to infer that a certain condition holds for the entire population. It involves setting up a null hypothesis that represents a baseline or no-effect scenario and an alternative hypothesis that represents the research question or effect being tested.

Null and Alternative Hypotheses

In hypothesis testing, the null hypothesis (H0) typically states that there is no difference or no effect, while the alternative hypothesis (H1) posits that there is a difference or an effect. The process involves assessing the evidence from the data to decide whether to reject H0 in favor of H1.

Chi-Square Test

The chi-square test is used for categorical data to determine whether there is a significant difference between the expected frequencies and the observed frequencies in one or more categories. In the context of survey data, it helps in testing whether the distribution of opinions differs from what would be expected under the null hypothesis.

p-value Approach

The p-value approach is a statistical method where the probability of obtaining the observed data, or something more extreme, is calculated under the assumption that the null hypothesis is true. If this probability, the p-value, is lower than a predetermined significance level (alpha), the null hypothesis is rejected.

Classical Approach

The classical approach involves comparing the calculated test statistic against a critical value derived from a relevant distribution (such as the chi-square distribution) corresponding to a chosen significance level. If the test statistic exceeds the critical value, the null hypothesis is rejected.

Significance Level

The significance level, often denoted by alpha, is a threshold set by the researcher (commonly 0.05) that determines the probability of rejecting the null hypothesis when it is actually true. It serves as the criterion for deciding whether the observed effect is statistically significant.

Transcript

00:01 This question asks us about the differences in staffing situations at magnet hospitals versus non -magnant hospitals.

00:08 A survey was given to a magnet hospital, and they responded about the staffing situation, and we want to see if the survey that we gave to a non -magnant hospital shows any sort of statistically significant difference in staffing situations.

00:24 What's going to be our null hypothesis? well, our null hypothesis is what would be easiest to believe.

00:30 It's easiest to believe that there is no difference in that all staffing situations are the same regardless of magnetness or not.

00:39 So to put that in numbers, we'd say that the probability of the first response in our survey to non -magnant hospital will be the same rate of the first response in our survey to a magnet hospital.

00:52 So p of 1 will be 12 % of 1.

00:56 Likewise, p of 2 will be the same as our survey results from the magnet hospital.

01:03 0 .32.

01:07 P of 3 will be 0 .38.

01:10 P of 4 will be 0 .12, and p of 5, that's a 5, will be 0 .06.

01:25 So these are our null proportions.

01:30 What we call null proportions, how we would expect the proportions to fall if the null hypothesis was true.

01:38 So we know that in our survey to a non -magnon hospital, we tested 500, we surveyed 500 nurses, so n is 500.

01:48 And these are the results we observe.

01:50 I'm going to make a little table with an observed column, and we'll start figuring out what our kai squared is.

01:57 We know this is a multinomial experiment, so we're going to have to find a kye squared statistic to test.

02:01 Our hypothesis.

02:04 So we observed that 165 nurses responded the first option, 140 responded the second, 125, the third, then 50 and 20.

02:20 So how do these compare what we would expect? well, our expected values are going to be the total number of respondents times our null probabilities.

02:28 So for the first response, it's going to be 500 times 12%.

02:35 To get what we would expect.

02:37 So 500 times .12 is actually just 60.

02:40 So we would expect 60 nurses to have responded option 1 in the survey.

02:46 Like the lies, we would have expected 160 for the second option, 190 for the third, 60 for the fourth and 30 for the fifth.

02:57 Now what we need to do is find our observed minus expected squared.

03:03 This is going to be part of our kai square test statistic.

03:05 So we need to find the difference and square it.

03:09 Our first one when we do that is 11 ,025.

03:15 It's a pretty big difference.

03:18 Next we get a difference of 20, squared is 400.

03:22 Moving along, we'll get a difference squared of 4 ,225, then a difference of 10, which is 100 when it's squared, and again, another difference of 10, so 100.

03:34 So these are our squared deviations.

03:38 And now we need to divide them by our expected values.

03:41 So we have o minus e squared over the expected values.

03:45 And when we divide by our expected values, we get something that looks like this.

03:49 Right away, we have 183 .75...

Please give Ace some feedback