Question
In a study of health behavior, a random sample of 500 subjects from a major city...
In a study of health behavior, a random sample of 500 subjects from a major city was surveyed. In this random sample, 310 responded that they had dinner mostly after 8 PM.
- At 5% level of significance, does the data support that more than 60% of the residents in this city had dinner after 8 PM? (Must state null and alternative hypotheses and use p-value to conclude the test. You can use Chi-square Goodness of fit test or Binomial test in SPSS or Single Sample Proportion Test in R Commander. However, if you use a statistical software, you must show the output from the software.)
Null hypothesis:
Alternative hypothesis:
Report p-value and use it to draw the conclusion:
[Put your statistical software output here to support your answers above!]
- Report a 95% confidence interval for estimating the percentage of residents in this city had dinner after 8 PM. (SPSS does not have an option for this question. R has a binconf() function that can find the confidence interval proportion using the formula you learned in this course. You must show software output if you used it to solve this problem. If you do not know how to use a software, you can do your own computing and show your work.) [Please report your answer in two forms, estimate plus and minus margin of error, and also the interval form.]
Estimate ± margin of error form: _______ ± _______
Interval form: ( , )
[If you used statistical software, put the software output here to support your answers above!]
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