## Frequency Tables and Measures of Central Tendencies

## Creating a Detailed Frequency Table

## Generating a Histogram for the Table

## 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.