Statistics - Computing Assignment - Assessment – Answer

Statistics Assignment

Case Scenario/ Task

For the variable Do they like the tv show

1. find

   =phat1=proportion of people that liked the tv show with ending 1

=phat2=proportion of people that liked the tv show with ending 2

=phat3=proportion of people that liked the tv show with ending 3

1. find

   phat1-phat2

phat2-phat3

1. Give a chart that compares the proportion of the 3 different tv show endings

Question 2

1. find

   =xbar1=the average amount people would pay for the dvd of the tv show with ending 1

=xbar2= the average amount people would pay for the dvd of the tv show with ending 2

=xbar3=the average amount people would pay for the dvd of the tv show with ending 3

1. find

   xbar1-xbar2

xbar2-xbar3

question 3

needs you to give complete descriptive statistics and histograms for the variable

how much you would pay

for each of the 3 different endings so you need to split the dataset into 3 different endings

Section 1

The random number taken into consideration is 27th randomly selected student

Section 2

The three proportions is given below

=phat1=proportion of people that liked the tvs how with ending 1 = 20/35 = 0.571

=phat2=proportion of people that liked the tv show with ending 2= 15/27 = 0.556

=phat3=proportion of people that liked the tv show with ending 3= 27/41 = 0.658

Section 3

The data taken into consideration is 27th randomly selected student. It is recorded that the proportion of people that liked the tv show with ending 1 is 0.571, proportion of people that liked the tv show with ending 2 is 0.556 and the proportion of people that liked the tv show with ending 3 is 0.658

Section 4

=phat1=proportion of people that liked the tvs how with ending 1 = 20/35 = 0.571

=phat2=proportion of people that liked the tv show with ending 2= 15/27 = 0.556

=phat3=proportion of people that liked the tv show with ending 3= 27/41 = 0.658

phat1-phat2= 0.571 – 0.556 = 0.016

phat2-phat3 = 0.741 – 0.656 = 0.034

Question 2

5. find

   = average amount people would pay for the dvd of the tv show with ending 1 = 5.704

= average amount people would pay for the dvd of the tv show with ending 2 = 5.99

= average amount people would pay for the dvd of the tv show with ending 3 = 7.215

5. find

   xbar1-xbar2 = 5.704 – 5.99 = -0.231

xbar2-xbar3 = 5.99 – 7.215 = -1.23

Question 3

The descriptive statistics is given below

Section 5

Quantitative and qualitative are two kinds of statistical techniques that are widely used in statistical researches to differentiate the data and statistical analysis. Quantitative methods are normally used to highlight the objective measurements using the data collected through various survey techniques (polls, telephone survey, questionnaires and surveys). On the other hand, qualitative methods are social research methods which are limited to the researcher views. In order to measure the deviation of the variable taken into consideration, range, standard deviation, lower quartile and upper quartile are used. Measures of dispersion are used to determine the distance between the individual values to the central value. On considering the correlation coefficient value, the best strategy for Karen is to increase his production during the high temperature seasons and it is observed that his product has higher demand during hot seasons.

Section 6

The two types of sampling techniques are

• Probability sampling technique, and
• Non probability sampling technique

The probability sampling method is classified into four major types, namely

• Simple Random Sampling method
• Stratified sampling method
• Systematic Sampling method
• Cluster sampling method

In simple random sampling technique, all samples have equal chance of being selected

In stratified random sampling technique, the population of N elements is divided into H subgroups and samples are taken at certain proportion from each subgroup called strata

In systematic sampling technique, samples are grouped with equal class width so that each class interval will have the sub group size. The first sample is randomly selected and the k – 1 samples are selected from the remaining subgroups automatically. For example, let us consider that Karen has 10 subgroups with samples of 10 in each subgroup. Suppose, he selects a sample 3 from the first sub group. Then the remaining 9 samples are selected in the order of 13, 23, 33, 43, 53, 63, 73, 83 and 93 respectively

Cluster sampling technique is used when the population seems to be homogeneous. In this sampling method, the population is divided into clusters and simple random sampling technique is used to generate the samples from each clusters

Section 7

When the sample size is large, then the mean of the sampling distribution follows normal approximately. Let us consider an example for sampling distribution. For the samples of size 25, the mean of the sampling distribution of means will fall close to the population mean, That is,

Sample mean of the sampling distribution of mean =

Standard deviation =

Since the sample size for each sample is small, we can expect the shape of the distribution of sample means for samples of size 25 to be skewed

For the sample of size 100, the mean of the sampling distribution of sample means is the same as the population mean, That is,

Sample mean of the sampling distribution of sample mean =

Standard deviation =

Since the sample size for each sample is large, we can expect the shape of the distribution of sample means for samples of size 100 to be normal

The difference in proportions and the difference in means are given below

phat1-phat2= 0.571 – 0.556 = 0.016

phat2-phat3 = 0.741 – 0.656 = 0.034

xbar1-xbar2 = 5.704 – 5.99 = -0.231

xbar2-xbar3 = 5.99 – 7.215 = -1.23

The independent sample t test is given below

Since the p – value of z test statistic is 0.6829 > 0.05, there is no sufficient evidence to conclude that there is a significant difference in mean values between ending 1 and ending 2

Since the p – value of z test statistic is 0.0684 > 0.05, there is no sufficient evidence to conclude that there is a significant difference in mean values between ending 2 and ending 3

Section 8

The previous sections showed how to compute and interpret the descriptive statistics and graphs for continuous and nominal data. We need to use different statistical test in testing the continuous and nominal data. Here, we used independent sample test to test the mean number of hours difference and two proportion z test to test the difference in proportion of two values. In addition, hypothesis testing procedures also used appropriately to determine the difference in two mean groups