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Statistics: Chi Squared Test

Q1: The data below (which have been ordered to facilitate analysis) present the blood serum cholesterol concentrations (x_i as mg/100 ml) of 45 women aged between 40 to 45 years.

137 148 155 156 162 166 167 168 168 174 175 178 178 179 179
179 182 186 187 188 189 190 190 191 192 193 193 194 198 199
201 203 207 207 207 211 212 214 218 218 219 222 225 229 255

Note:
Sum of x_i = 8589; Sum of x_i^2 = 1663603

(a) Assuming that data are drawn from a normal population, find a 95% con fidence interval for the true mean of this population.

(b) Construct a histogram for the data, taking as grouping intervals; (- infinity; 160], (160; 180], (180; 200], (200; 220], and (220;infinity). Determine the expected number of observations in each interval of the histogram, for a sample size 45 drawn from a normal distribution having the mean and standard deviation computed in part (a)

(c) Hence test whether a normal distribution appears to be a reasonable model for these data.

Q2: A set of three dice were tossed 180 times, and on each occasion the number of faces showing a 5 or 6 was recorded. On 9 occasions three faces showed a 5 or 6; on 47 occasions two faces showed a 5 or 6; on 75 occasions only one face showed a 5 or 6; and on the remaining occasions no 5 or 6 was obtained. Are these data consistent with the three dice being unbiased?

Q3: In a clinical trial to assess the value of a new method of treatment (A) against the old method (B), patients were divided at random into two groups. Of 257 patients treated by method A, 41 died; of 244 patients treated by method B, 64 died. Test whether the death rates can be considered the same for the two treatments (again use both a 2-test, and an appropriate Normal approximation). Finally, find an approximate 95% con fidence interval
for the difference between the death rates for the two treatments.

Solution Preview

Q1
Firstly we will find the sample mean and the sample standard deviation :

= = = 36,969.02- 36,431.36 = 537.66
Therefore =23.19
a) For 95% CI of the sample mean, we have and
CI is: = 190.87 ± *1.96
= 190.87±6.77
= [184.10, 197.64]
b) I did this part in EXCEl using countif function and got:

Intervals Count
(-00,160] 4
(160, 180] 12
(180,200] 14
(200, 220] 11
(220, 00) 4
Total 45

In part a) we got Mean =190 and st. dev = 6.77
Using the 68-95-97.5 rule for normal distribution, there should be about 68% of the data within one standard deviation from the mean, 95% within 2 st. dev from the mean and 97.5% within 3 st dev from the ...

Solution Summary

Hypotheses Testing using Chi-square test. Constructing 95% Confidence Interval for the true mean of the population.
Finding Central tendency estimators.

$2.19