SPSS Analysis

Question 2

Part b

The n or the sample is more than 30 and therefore considered to be enough for inferring the results for the whole population

The observations doesn't appear to be in a pattern, therefore the criteria of random selection, tends to be fulfilled.

Last and the most important assumption relates to the approximate normality of the population distribution, in terms of n.

As it can be seen in the graph above, most of the values lie in the middle, near the values of the mean and the median, therefore we can say that the population is approximately, normally distributed.

Part C

In order to calculate the 95 % confidence interval, we shall first calculate the standard error, which can be calculated as follows:

s/vN = 38.36 / v142

= 3.22

Now we can put in the value of Standard Error, multiply it with the Z score at 0.025 i.e. ±1.96 and add / subtract it with the mean, in order to get the 95 % confidence interval.

150.88 - 1.96 (3.22) = µ = 150.88 + 1.96 (3.22)

144.5 = µ = 157.2

Part D

H0: µ = 140

H1: µ > 140

Where µ stands for the mean value to be tested, i.e. 140

The value of the suitable test statistics equals to

Z= X - µ / (s/vn)

Z = 150.88 - 140 / (38.36/v142)

Z = 3.38

The p value of this test is 0.001(Correct to 3 decimal places, as seen in the Z score table)

At the 5% or 0.05 significance level, if we evaluate our hypotheses, we can conclude that out null hypothesis, i.e. H0: µ = 140, will be rejected because the p value < 0.05 which means that mean Systolic Blood Pressure for diabetics is > 140.

Part E

One-Sample Statistics

N

Mean

Std. Deviation

Std. Error Mean

sysBP

142

150.88

38.361

3.219

History of Diabetes ...