Final answer:
To perform a Chi-Square test for the hotel patron preferences, calculate the expected counts for each hotel aspect, input both observed and expected counts in Excel, and apply the CHISQ.TEST() function to obtain a p-value. A low p-value indicates a significant difference from the expected distribution, while a high p-value suggests the observed distribution fits the null hypothesis.
Step-by-step explanation:
To test the hypothesis that people are equally likely to select any one of the five aspects they value most in a hotel (cleanliness, amenities, service, location, and price), we first need to calculate the expected counts for each category. With 300 patrons surveyed, and 5 categories, the expected count for each category if the null hypothesis is true (that is, that each category is equally likely) would be 300 patrons divided by 5 categories, resulting in 60 expected responses per category.
Now using Excel to calculate the Chi-Square test statistic, we would enter the observed counts (70, 60, 45, 55, 70) and the expected counts (60, 60, 60, 60, 60) into two separate columns. We would then use the CHISQ.TEST() function in Excel by entering the range of observed counts and expected counts as the arguments to the function. This will return the p-value for the Chi-Square test.
If the p-value is low (typically below 0.05), it would suggest that there is a significant difference between the observed and expected counts, meaning the null hypothesis that people are equally likely to choose each of the five values in a hotel is rejected. Conversely, a high p-value implies that the observed distribution of responses fits well with the expected distribution under the null hypothesis.