Question

Use the Roll Verification Data 1) Create a Histogram for the Height data. Convert to feet only to do this.ex 5 feet 3inches is 5+3/12=5.25 2) Calculate the mean, median, standard deviation A Hypothesis test 3) Jed raises chickens. He says that the average chicken weight is 5.5 lbs. Test his claim. A simple random sample, looks like it's from a Normal distribution of Jed's chickens was collected with the following data in pounds: 3.4, 5.8, 6.0.4.2.5.7.7.1.6.3, 6.2.5.1,7.2 (you find the stats)

Answer 1

1)

> hist(mydata$Height,col = 'Blue')

Histogram of mydata $Height 8 9 Frequency 4.8 5.0 5.2 5.4 5.6 5.8 6.0 mydata $Height

2)

> mean(mydata$Height)
Mean = 5.420192
> median(mydata$Height)
Median= 5.333333
> sd(mydata$Height)
Standard Deviation= 0.3134412

3)

Hypothesis:

H₀: $\mu_0 = 5.5$

H_a: $\mu_0 \neq 5.5$

As significance level = 0.05(default)

Critical value

As we dont have the population standard deviation, we need to perform 1 sample t test.

t critical at $\alpha$ = 0.05 is 2.262156

If t stat calculated is greater than t critical , we can reject the null hypothesis.

Test statistic

	Weight
	3.4
	5.8
	6
	4.2
	5.7
	7.1
	6.3
	6.2
	5.1
	7.2
Mean($\overline{X}$)	5.7
Standard Deviation(S)	1.193501

t stat=$\frac{\overline{X}- \mu_0}{s/\sqrt{n}}$=  $\frac{5.7- 5.5}{1.1935/\sqrt{10}}$

= 0.5299

Result and Interpretation

As t stat(0.5299) < t critical (2.262156) , we cant reject the null hypothesis.

Therefore we can conclude that there is no sufficient evidence to say that the mean weight of chicken is different from 5.5lbs.