Question

An internet company guarantees on average a 60 Mbytes/s download speed or higher. You are downloading a new engineering software package and it seems to be taking a long time. You decide to test the internet company’s claims and run a download speed test at 10 different times, resulting in the following measurements: 59, 45, 70, 33, 44, 19, 75, 60, 55, 58 (all units in Mbyte/s). Use α = 0.05.

1. Follow the 7 steps for hypothesis testing to solve this problem using the critical region method.

2. Calculate the p value.

3. Calculate the 95% confidence interval. Look to see if the hypothesized mean is in confidence intervals.

4. Do you come to the same conclusion using critical region, p value and confidence intervals?

do the below

include lower and upper confidence intervals

Identify parameter of interest

Formulate null hypothesis Ho

Formulate alternative hypothesis H1

State appropriate test statistic

Choose level of significance (α) and determine critical region

Perform computations

Make and state conclusion

Answer 1

	Values ( X )
	59	51.84
	45	46.24
	70	331.24
	33	353.44
	44	60.84
	19	1075.84
	75	538.24
	60	67.24
	55	10.24
	58	38.44
Total	518	2573.6

X 518/10= 51.8

Standard deviation S_{X} =

Sx = V25736/10-1 = 16.9102

To Test :-

H0 :-  

μ < 60

H1 :-

>60

Test Statistic :-


t (51.8-60)/(16.9102/V10)

t = -1.5334

Test Criteria :-
Reject null hypothesis if
t > ta,n-1


a,n-1 -to.o5,10-11.833 an-I


t>tan-1-1 1.53341.833

Result :- Fail to reject null hypothesis

Part 2)

P value

Looking for the value t = 1.5334 across n - 1 = 10 - 1 = 9 degree of freedom in t table

t = 1.5334 lies between the value 1.383 and 1.833 with respective P value 0.10 and 0.05

P value = 0.0798

Reject null hypothesis if P value < level of significance

0.0798 > 0.05, fail to reject null hypothesis

Conclusion :- We Accept null hypothesis

There is no sufficient evidence to support the claim that an internet company guarantees on average a 60 Mbytes/s download speed or higher.

Confidence Interval




Lower Limit = 39.7032

Upper Limit = 63.8968
95% Confidence interval is ( 39.7032 , 63.8968 )

Since  $mu=60$ lies in the interval  ( 39.7032 , 63.8968 ) , hence we can conclude to accept null hypothesis