health and physical condition
Part I
Please use the sample data from students to perform the following tasks. All subjects in the sample are in normal health and physical condition.
Link to Data: http://gchang.people.ysu.edu/stat/ClassdataMPH_19x.xls
(The weight data are in pounds and the height data are in inches.)
(Use weight and height data to compute Body Mass Index (BMI). Please find the BMI Formula from Internet and compute the compute BMI using statistical software.)
⦁ Test whether the average Body Mass Index for the student population where the student sample was drawn is different from 25, at 5% level of significance, using the sample data above. You must state null and alternative hypothesis, check normality assumption, report p-value, test statistic value, and draw a proper conclusion.
Null hypothesis: Mean = 25, The average BMI is 25.
Alternative hypothesis: The average BMI is different from 25.
Report p-value from the normality test and conclusion: P-value = 0.436, since p value is bow less than 0.05, so we do not reject null hypothesis.
Report p-value from the t-test and the conclusion: P-value = 0.95. Since the p-value is not less than 0.05, so we do not reject null hypothesis. There is no sufficient evidence to support the alterative hypothesis that the average BMI is statistically significantly different from 25.
⦁ Find the sample size so that the test can have a 95% power to detect a BMI average that is half a standard deviation from the mean, 25, (25 is the mean under the null hypothesis) at 5% level of significance, using a two-sided test.
⦁ Find the 95% confidence interval for estimating the average BMI for the sampled population.
(23.3487, 26.5517) 24.9802±1.6015
Descriptives
Statistic Std. Error
BMI Mean 24.9502 .78965
95% Confidence Interval for Mean Lower Bound 23.3487
Upper Bound 26.5517
5% Trimmed Mean 24.7723
Median 25.5266
Variance 23.072
Std. Deviation 4.80328
Minimum 15.62
Maximum 36.99
Range 21.37
Interquartile Range 6.63
Skewness .404 .388
Kurtosis .408 .759
⦁ Find the sample size that would be required if you wish the estimation of the mean to have a margin of error no more than 0.2. (Use the data to estimate the standard deviation for estimating the sample size.)
N= = 1.96² * 2.82 / 0.22 = 753
Mean
Confidence Level 0.95
2.8
Error (e) 0.2
Z 1.959964
Sample Size 752.9259
Sample Required 753
Part II
In a study of health behavior, a random sample of 400 subjects from a major city was surveyed. In this random sample, 250 responded that they had dinner mostly after 8 PM. (You need to use Weight Cases option to generate the data in SPSS data editor.)
⦁ At 5% level of significance, does the data support that percentage of the residents in this city had dinner after 8 PM is different from 60%? (Must state null and alternative hypotheses and use p-value from a z-test to conclude the test. You can do this problem by hand or use the Excel Worksheet for One Sample Inference posted on Blackboard.)
Null hypothesis: p =.6; percentage of people who had dinner after 8 pm is 60%
Alternative hypothesis: p .6; percentage of people who had dinner after 8 pm is not 60%
Report p-value from the z-test and the conclusion:
P value is larger so it is not less than .05 so we fail to reject the null hyopothesis there is no sufficient evidence for the alertaive of the percentage.
⦁ Perform a Chi-square Goodness of Fit test to test the proportion for part a) using SPSS and report the p-value.
[Place your software output here.]
⦁ Report the 95% confidence interval to estimate the percentage of people in the sampled population who had dinner mostly after 8 PM using either Wald’s normal approximation method, Wilson’s method, or Clopper-Pearson’s method using SPSS. Must use interval notation to report the confidence interval and comment on which method was used.
(57.756%, 67.244%)
62.5± 4.744%
