Question

According to the US Government 68% of adults weigh more than they should. A dietician, Dr. Scales, believes that drinking 8 or more glasses of water a day was associated with not being overweight. She obtained a random sample of 580 American adults who consumed 8 or more glasses of water a day, weighed each adult, and classified each one as overweight or not. The table below shows the observed frequencies and the expected frequencies. Use alpha = .05.

	Not OVerweight	Overweight
Observed	230	350
Expected	185.60	394.40

a. Write the null hypothesis.

b. Write the alternative hypothesis.

c. What type of test would be used to analyze this data? Be specific.

d. What are the percentages of overweight and not overweight participants in the study?

e. What is the critical value for alpha = .05? (need the table in the book)

f. Is the result of the statistical test significant?

g. Write the result in correct APA statistical notation.

h. Summarize the results in one or two sentences.

Answer 1

We have an assumption that 68% of the people are over wieght which forms the population and its parameters , and Dr scales believes that those who drink 8 glass a day are not over wieght and he is trying to prove it through 580 samples who drink 8 glass of water a day , To say about the claim of Dr. Scales, we will prove only one thing is there any difference between the sample and the population. No difference means Dr. scales assumption is wrong and there is diference means Dr Scales is right

a) The null Hyporhesis Ho: There iso significance difference between the observed frequnecy and the expected frequency

b) The aternate hypothesis H1: There is significance difference between the observed frequency and the expected frequency

c) If we have given the observed frquency and expected frequency and we need to test the goodness of fit (means whetherr the sample fit to the population) The best test will be Chi square test.and the test statistic will be

where Oi is the observed frequnecy and Ei is the expected frequency of corresponding data.

e) Using the above formula we need to find the Chi square value ( shown in the below table

We have the critical vaue value of 15.62 we have given that the significan level = .05 and degree of freedom will be 2-1 as there are only two values we have 2-1 = 1

f) The p value of the above from the p value calculator we have (using above details) p = 0.000077 which is < 0.05 which is the significance level .

We know that if the p value is less than the significance level we declare that the statistical result is sigificant.

g) The correct statistical APA notation for the test is

= 15.62

degree of freedome df= 1

Significance level 0.05

P value = 0.000077<

h) here we have the p value very smaller than the significance level Alfa, which means that there is no way we can accept the null hypothesis , means we have to reject the null hypothesis , which naturally says that there is some significant differnce between the observed values and the expected values expection was 68% over weight but the fact is different with the sample of people taking 8 glasses of water a day , This says Dr. Scales inter pretation has some truth in it