# SPSS Data Analysis Homework Solution

Get access to free solutions on SPSS data analysis online on our website. The solutions developed by us are flawless and straight to the point. That's why you can use them to learn and revise for exams. What's better, we don't charge you a dime to go through them. Get access to free solutions on SPSS data analysis online on our website. The solutions developed by us are flawless and straight to the point. That's why you can use them to learn and revise for exams. What's better, we don't charge you a dime to go through them.

## Frequency Tables and Measures of Central Tendencies

Below are solutions to questions on frequency tables, measures of central tendencies, and other concepts of data analysis using SPSS. Mean, median, and standard deviation are some of the calculated parameters. A histogram pie chart, dot plot, and frequency polygon are constructed to give a visual presentation of the data.

## Creating a Detailed Frequency Table

Description: The data set represents the number of minutes a sample of 27 people exercise each week

108 139 120 123 120 132 123 131 131

157 150 124 111 101 135 119 1 6 117

127 128 139 119 118 114 127 142 130

1.- Construct a frequency distribution for the data set. Include class limits, midpoints boundaries, frequencies, and cumulative frequency. Describe each graph. ## Generating a Histogram for the Table

2.- Display the data, using a frequency histogram and a frequency polygon.  3.- Display the data using a stem and leaf plot. 5.- Make a pie chart graph. ## Constructing a Dot Plot ## Finding the Measures of Central Tendency Mean =∑▒〖∑▒fx/f=3421/27=126.7037〗

Median = 1/2 (27)=13.5

Therefore, the median will be in the class limit 113-125

Mode = 11 which is the range for 113-125

Standard deviation = √(∑▒(f(X-M)^2)/f)=√(4481.63/27)=12.8836

## Finding the coefficient of variation and CI

Coefficient of variation =σ/μ=12.8836/126.7036=0.1017=10.17%

Confidence interval for Mean

confidence interval=x ̅±Z σ/√n

=126.7037±1.96(12.8836/√27)

=12.7037±4.8597

=(121.844,131.536)

The confidence interval for the mean is 122.844 and 131.536

## Testing the hypothesis

10.- The coach claims that meantime the number of 27 people exercising is more than125 minutes each week. Can you support the claim, using a level of significance of 0.05?

Ho:μ=125

H1:μ>125

Level of significance =5% = 0.05

Test statistics =(x-μ)/(σ/√n)=(125-126.7036)/(12.8836/√27)=-1.7037/2.4794=-0.6871

Test statistic =|z|=0.6871

Critical value =Z_(α/2)=1.645

Conclusion: Since the test statistic is less than the critical, we fail to reject the null hypothesis because there is not enough evidence to support the claim that the meantime of 27 people exercising is more than 125 minutes each week.