Solution Code: 1AEIC
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Here, we wish to determine whether the mean amount spent for snacks differ significantly between the male and female respondents. Recent days, people tend to spend more money on snacks instead of healthy foods. Foods such as pizza, burger, chips, and high sugar contained candies are sold at higher rates which the affects the health of consumers and they are not aware of the root cause of their health disorders. Fast foods contain high amount of calories and very little amount of nutrition. There will be rapid increase in weight if a person consumes fast food items continuous and he will face severe health disorders. This study is carried out mainly to create an awareness over the people, and for this purpose, a random sample of 100 respondent’s opinion was taken regarding the amount spent for their snack food.
Description of the dataset
The variable taken into consideration is given below
Which version is the best ? Ordinal Variable (Neither, Version 1 and Version 3)
Gender ? Nominal (Male and Female)
Do you like the product ? Ordinal (Like and Hate)
How much would pay? ? Interval
Are they old? ? Interval
Summary of the dataset
Minimum: | 0 |
Maximum: | 3.3 |
Range: | 3.3 |
Count: | 100 |
Sum: | 251.9 |
Mean: | 2.52 |
Median: | 3.1 |
Mode: | 3.1 |
Standard Deviation: | 1.21 |
Variance: | 1.47 |
Mid Range: | 1.65 |
Quartiles: | Quartiles:
Q1--> 0.5 Q2--> 3.1 Q3--> 3.2 |
Regarding the distribution of the amount spent on snack food, it is observed that the distribution of amount spent on snack food has longer tail towards the left side, indicating that the distribution of amount spent on snack food is skewed left. This means that mean amount spent on snack food (2.598 ± 1.16) is significantly less than that of median amount spent on snack food (median = 3.1The minimum and maximum recorded amount spent on snack food is 0 and 3.3
Bi)
Row Labels | Count of Which version is the best? |
neither | 11 |
version 1 | 26 |
version 2 | 41 |
Grand Total | 78 |
So proportion of neither =11/78=O.14,
Proportion of version 1 = 26/78=O.33,
Proportion that like version2 = 41/78 =O.53
Regarding the respondents preference over their product, about 53% prefer for version 3 and 33% prefer for version 1 product. This indicates that, respondents prefer to Version 3 ahead to Version 1
3bii)
Minimum: | 3 |
Maximum: | 3.3 |
Range: | 0.3 |
Count: | 78 |
Sum: | 246.1 |
Mean: | 3.16 |
Median: | 3.2 |
Mode: | 3.1 |
Standard Deviation: | 0.108 |
Variance: | 0.0116 |
Mid Range: | 3.15 |
Quartiles: | Quartiles:
Q1--> 3.1 Q2--> 3.2 Q3--> 3.2 |
Regarding the distribution of the amount spent on snack food the consumers like, it is observed that the distribution of amount spent on snack food has longer tail towards the left side, indicating that the distribution of amount spent on snack food is skewed left. This means that mean amount spent on snack food they like (3.16 ± 0.108) is significantly less than that of median amount spent on snack food (median = 3.2). The minimum and maximum recorded amount spent on snack food is 3 and 3.3
3c Foreach pair of variables below give appropriate graphical and numerical summaries that describes the variable (in other words give the appropriate bivariate statistics and graphical displays)
i)The variables: Gender and do they like the product ?
ii)The variables How much they would pay?andgender
3) ci
Proportion of people that like the product | Column Labels | ||
Row Labels | like | hate | Grand Total |
male | 77.27% | 22.73% | 100.00% |
female | 78.57% | 21.43% | 100.00% |
Grand Total | 78.00% | 22.00% | 100.00% |
count of people that like product | Column Labels | ||
Row Labels | like | hate | Grand Total |
male | 34 | 10 | 44 |
female | 44 | 12 | 56 |
Grand Total | 78 | 22 | 100 |
Regarding the association between gender and products they like, about 77.27% of male respondents and 78.57% of female respondents said that they like the products. Thus, there is no association between gender and product they like
3cii)
For the MALES the summary of the amount they would pay is
Minimum: | 0 |
Maximum: | 3.3 |
Range: | 3.3 |
Count: | 56 |
Sum: | 141.8 |
Mean: | 2.53 |
Median: | 3.1 |
Mode: | 3.1 |
Standard Deviation: | 1.18 |
Variance: | 1.4 |
Mid Range: | 1.65 |
Quartiles: | Quartiles:
Q1--> 0.5 Q2--> 3.1 Q3--> 3.2 |
For the FEMALES the summary of the amount they would pay
Minimum: | 0 |
Maximum: | 3.3 |
Range: | 3.3 |
Count: | 44 |
Sum: | 110.1 |
Mean: | 2.5 |
Median: | 3.1 |
Mode: | 3.2, 3.3 |
Standard Deviation: | 1.26 |
Variance: | 1.58 |
Mid Range: | 1.65 |
Quartiles: | Quartiles:
Q1--> 0.3 Q2--> 3.1 Q3--> 3.2 |
Regarding the distribution of the amount spent by male respondents on snack food, it is observed that the distribution of amount spent on snack food has longer tail towards the left side, indicating that the distribution of amount spent on snack food is skewed left. This means that mean amount spent on snack food (2.53 ± 1.18) is significantly less than that of median amount spent on snack food (median = 3.1). The minimum and maximum recorded amount spent on snack food is 0 and 3.3. Since the distribution is skewed, we can use Median for measuring the central tendency
Regarding the distribution of the amount spent by female respondents on snack food, it is observed that the distribution of amount spent on snack food has longer tail towards the left side, indicating that the distribution of amount spent on snack food is skewed left. This means that mean amount spent on snack food (2.5 ± 1.26) is significantly less than that of median amount spent on snack food (median = 3.1). The minimum and maximum recorded amount spent on snack food is 0 and 3.3. Since the distribution is skewed, we can use Median for measuring the central tendency. Here, we have two modal values and therefore, the distribution is bimodal.
Going through the histogram, we see that the highest amount spent on snack food is female respondents and they spend slightly higher than that of their male counterparts
Confidence Intervals
p-Z2*p1-pn,p+Z2*p1-pn
=0.33-1.645*0.331-0.6778,0.33+1.645*0.331-0.6778
= (0.2455, 0.4211)
The 90% confidence interval for the proportion of people that prefer version 1 is (0.2455, 0.4211). This indicates that, when more number of samples are taken from the same population, then there is a 90% (90 out of 100 times) chance that the true proportion of people that prefer version 1
ii)
The 90% confidence interval for the average amount they would pay is calculated by using the formula given below
x-Z2*n,x+Z2*n=2.598-1.645*1.16100,2.598+1.645*1.16100
= (2.32, 2.72)
The 90% confidence interval for the average amount they would pay is (2.32, 2.72). This indicates that, when more number of samples are taken from the same population, then there is a 90% (90 out of 100 times) chance that the true average amount they would pay will fall between 2.32 and 2.72
Hypothesis Testing
In order to determine whether there is an association between gender and product preference (like or hate), we perform chi – square test for independence. The null and alternate hypotheses are given below
Null Hypothesis: H0:
That is, there is no association between gender and product preference (like or hate)
Alternate Hypothesis: Ha:
That is, there is an association between gender and product preference (like or hate)
Level of Significance: The level of significance is set to ? = 0.05
Test Statistic
The chi – square test statistic is
2=Oi-Ei2Ei
The table given below shows the workings of Chi – Square test statistic
Sl. No | Observed Frequency (Oi) | Expected Frequency (Ei) | (Oi - Ei)^2 | (Oi - Ei)^2/Ei |
1 | 34 | 34.32 | 0.1024 | 0.002984 |
2 | 44 | 43.68 | 0.1024 | 0.002344 |
3 | 10 | 9.68 | 0.1024 | 0.010579 |
4 | 12 | 12.32 | 0.1024 | 0.008312 |
Total | 100 | 100 | Chi Square test statistic | 0.0242 |
P - value | 0.8763 |
From the above table, it is observed that there is no association between Gender and do they like the product (Chi – Square test statistic = 0.0242, p – value = 0.8763 > 0.05)
ii)
In order to determine whether there is a difference between the mean amount males would pay and the amount females would pay, we perform independent sample t test. The null and alternate hypotheses are given below
Null Hypothesis: H0: µ1 = µ2
That is, there is no difference between the mean amount males would pay and the amount females would pay
Alternate Hypothesis: H0: µ1 ? µ2
That is, there is a difference between the mean amount males would pay and the amount females would pay
Level of Significance: The level of significance is set to ? = 0.05
The table given bellow shows the workings of independent sample t test
Data | Confidence Interval Estimate | |||
Hypothesized Difference | 0 | for the Difference Between Two Means | ||
Level of Significance | 0.05 | |||
Female Pay | Data | |||
Sample Size | 56 | Confidence Level | 95% | |
Sample Mean | 2.532143 | |||
Sample Standard Deviation | 1.183386 | Intermediate Calculations | ||
Male Pay | Degrees of Freedom | 98 | ||
Sample Size | 44 | t Value | 1.9845 | |
Sample Mean | 2.502273 | Interval Half Width | 0.4864 | |
Sample Standard Deviation | 1.257811 | |||
Confidence Interval | ||||
Intermediate Calculations | Interval Lower Limit | -0.4565 | ||
Population 1 Sample Degrees of Freedom | 55 | Interval Upper Limit | 0.5162 | |
Population 2 Sample Degrees of Freedom | 43 | |||
Total Degrees of Freedom | 98 | |||
Pooled Variance | 1.4801 | |||
Standard Error | 0.2451 | |||
Difference in Sample Means | 0.0299 | |||
t Test Statistic | 0.1219 | |||
Two-Tail Test | ||||
Lower Critical Value | -1.9845 | |||
Upper Critical Value | 1.9845 | |||
p-Value | 0.9032 | |||
Do not reject the null hypothesis |
From the above table, we see that the value of t test statistic is -0.4376 and its corresponding p – value is 0.663 > 0.05. This indicates that there is no difference between the mean amount males would pay and the amount females would pay
Problems of getting survey data in the real world
Online survey data, is the recent trend in data collection technique, as it becomes difficult to collect the data by meeting the respondent directly. Since the respondents are not interested in spending time for the survey when they are visited directly as they feel worry about the confidentiality issues and beneficial issues, direct survey technique seems to be less favored when compared online survey technique. Also, the cost and time consumption is more when the data is collected in real world. The major drawback in getting the survey data in the real world is the accuracy of the information, as respondents who are busy in their personal and official commitments, normally tend to finish the survey by either giving false or incomplete information
Conclusion
The study main objective is to create an awareness to the respondents on the health defects of eating more snack foods. For this purpose, a random sample of 100 respondents detail about the amount they spent on snack food and other various information are collected. It is found that female respondents, generally spend more on snack food when compared with male counterparts. In addition, there is no association between gender and the product they prefer. Overall, the mean amount spent on snack food is found to be 2.598 with a standard deviation of 1.16 and the median amount spent is 3.1. The recorded least amount spent on snack food is 0 and the highest amount spent on snack food is 3.3. Here, we see that the respondents spend more amount on snack foods which will lead to various health hazards and obesity. Therefore, proper awareness camp must be carried out to create an awareness over the ill effects of consuming snack food
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