Data Analysis



Data Analysis

Probability

Question 1

Probability of success = 625/780

p = 0.801,q = 1-p = 0.198

Mean = = np = 625

Standard deviation = = 780 625 0.198 = 124.2

Now, for the 98% confidence interval

= 625 3.250(124.2)

= (221.35, 1028.65) is the 98% confidence interval for the households having personal computer.

Yes, the above procedure contains all the assumptions and working for the 'rule of thumb'.

Using the formula n = = = 4251.69 4252 units.

Question 2

The mean and standard deviation using SPSS is given below:

Descriptive Statistics for Gestation

Mean

Std. Deviation

N

Gestation

277.464

15.7565

218

The conditions for finding a confidence interval or checking a hypothesis contains the sample mean (), population mean (x-bar), standard deviation of the sample () and total observations (n). They all can be easily found with the given data.

CI = Mean (SD) = 277.46 1.29 (15.757) = (257.133, 297.786) is the 90% confidence interval for the mothers who are over 30 years of age.



Ho : 280 days

H1: 280 days

Zcal = = = 2.39

P (Z < 2.39) = 1 - 0.9916 = 0.0084.

The resulting p - value is highly significant because it is less than 0.05, therefore we reject the null hypothesis and can claim that there is no evidence that the gestational period is less than 280 days for mothers who are over 30 years in age.

Question 3



Descriptive Statistics for Gestation

Mean

Std. Deviation

N

Birth weight (grams) for previous pregnancy

3385.16

491.9175

200

Birth weight (grams) for no pregnancy

3346.406

476.7603

170



Let the average birth rate for no pregnancy is and for 1 pregnancy is.

H0: =

H1:

The assumptions for finding the independent t - statistic contains the mean of no pregnancy and the mean of 1 pregnancy which is available and also the combined standard deviation can also be found out with the help of the formula. All the assumptions are available for carrying out this test.



Now, Sx1-x2= = = 50.4675

t = = = -0.411

P (Z < -0.411) = 1 - 0.318 = 0.682.

The results of the p - value showed that the average birth weight of babies for mothers who have had one previous pregnancy differ from mothers who have had no previous pregnancies are same because P>0.05 and we cannot reject the hypothesis.

Question 4

The sampling distribution of a mean of a given statistic depends on the random sample. They have sheer importance in statistics because the major specification for the route to get the statistical inference can be easily got through it.

The major importance of normal distribution is ...