Question

Mayr et al. (1994) took an SRS of 240 children who visited their pediatric outpatient clinic. They found the following frequency distribution for the age (in months) of free (unassisted) walking among the children: 11 9 10 12 13 14 15 16 17 18 19 20 Age (months) Number of Children 13 35 4469 36 24 7 3251 64 Chapter 2: Simple Probability Samples Construct a histogram of the distribution of age at walking. Is the shape normally distributed? Do you think the sampling distribution of the sample average will be normally distributed? Why, or why not? a b Find the mean, SE, and a 95% CI for the average age for onset of free walking. Suppose the researchers wanted to do another study in a different region, and wanted a 95% CI for the mean age of onset of walking to have margin of error 0.5. Using the estimated standard deviation for these data, what sample size would they need to take? c

Answer 1

Ame. Date: ----21/5/20159 a) The objective is to construct the histogram of the distribution of age at walking Using Excel, follow the steps below to obtain the histogram for the age at walking data. 1) Select the range of the cells containing the Number of Children data. 2) Click on Insert>Histogram icon 3) Specify the range of the cells containing the values for the X-variable. ) Click on OK Output: 80 7 70 60 s 40 2 30 3 20 10 0 9 10 11 12 13 14 15 16 1718 19 20 Age

From the above histogram, it can be observed that the data spread more towards to the right So, the distribution of the data is positively skewed Therefore, the distribution of age at walking is not normal, the average value for the variable age wise number of children does not follow normal distribution

b) The objective is to find the mean, standard error and 95% confidence intervals for the average for onset of free walking Consider the calculations shown in the below table: 9 13 117 1053 10 35 350 3500 1144 484 5324 12 69 828 9936 13 36 468 6084

14 24 336 4704 15 7 105 1575 16 3 48 768 17 2 34 578 185 90 19 1 19 361 201 20 400

Determine the mean value using the following formula 117+350+... +19+20 13+35 +44+...+5+1+1 2,899 240 - 12.0792 Determine the standard deviation using the following formula SD 35,903_(12.0792) 240 1.9206 Determine the standard error using the following formula 1.9206 /240 -0.124

The 95% confidence interval for the average age for onset of free walking is shown below: x-a/2-1) x (SE), +t 12.0792 -)(0.124), 12.0792+(00/2.-)x(0.124)) (12.0792-1.96x(0.124), 12.0792+1.96x(0.124) (12.0792+0.24304) (11.83616, 12.32224) (11.8362, 12.3222) Hence the 95% confidence interval for the average age for onset of free walking is lies between 11.8362, 12.3226

c) The objective is to determine the sample size. It is given that, margin of error, e 0.5 Level of the significance, α-0.05 The estimated standard deviation, S-1.9206 From the normal distribution tables, the critical value at 5% level of the significance about two tailed test is 1.96 Determine the sample size using the following formula: 2 (1.96x 1.92062 0.5 56.68 e 57

Then the desired sample size is, 7 57 57 1+ 240 57 1 + 0.2375 57 1.2375 = 46.06061 47 147. Therefore, the sample size is