Question

2. Battery Level of Laptop Computers in Shipment A computer manufacturer ships laptop computers with the batteries fully charged so that customers can begin to use their purchases right out of the box. In its last model, 85% of customers received fully charged batteries. To simulate arrivals, the company shipped 100 new model laptops (randomly picked from their warehouse) to various company sites around the country. Of 100 laptops shipped, 96 of them arrived reading 100% charged. Do the data provide evidence that the proportion of new model laptop computers arrived fully charged to various company sites around the country of a computer manufacturer is higher than the last model? Test an appropriate hypothesis at α = 0.05? Use RStudio.

1. Label the parameter: (4 points)

2. State null and alternative hypotheses: (6 points)

H₀:

H_a:

3. Propose an appropriate hypothesis test: (4 points)

4. Verify the required conditions for the proposed hypothesis test: (8 points)

Randomization assumption:

Normality assumption:

5. Find test statistic and p-value: (9 points)

Test statistic = 

p-value = 

Note: For your future reference, save your R codes and outputs on your machine:

6. Interpret test statistic: (4 points)

7. Interpret the P-value in context: (6 points)

8. Make a decision of hypothesis test: (4 points)

1. Make a conclusion of hypothesis test in context: (5 points)

Answer 1

Here population parameter of interest is population proportion (p)

we have to test is p is 0.85 or greater than 0.85

So H0 : p = 0.85

vs Ha : p >0.85

so this is right tailed proportion test so here we have to use z-test statistics as follows

z = (phat - p)/ sqrt( p*q/n) where phat = x/n = 96/100 =0.96

z = (0.96-0.85) / sqrt( 0.85*0.15/100) = 3.08

Here we assumed that our population follows normal distribution as np = 100* 0.85 = 85 >=10

and nq= 100*0.15 = 15 >=10

also we assumed that The data are a simple random sample from the population of interest.

so here test statistics z = 3.08

p-value = Proba.( Z > 3.08) = 1 - NORMSDIST(3.08) =1 - 0.998965=0.001035

here p-value is 0.001 which is less than 0.05 so we have to reject H0.

means we are not rejecting Ha.So we conclude that p>0.85.

So we can say that proportion of new model laptop computers arrived fully charged to various company sites around the country of a computer manufacturer is higher than the last model.