A manufacturer wants to inspect the absorption capacity of a new sponge. Based on past data, the average sponge could absorb 3.5 ounces. After the redesign, the absorption amounts of a sample of 10 sponges were (in ounces): 4.1, 3.7, 3.3, 3.5, 3.8, 3.9, 3.6, 3.8, 4.0, and 3.9. At a level of significance of 1%, would you claim that the new design is better than the old one?

Requried:
a. No because we don't reject the Null Hypothesis at Alpha = 0.01
b. Yes because we reject the Null hypothesis at Type 1 error = 1%
c. Yes because the cut-off is more than the sample mean
d. No because the cut-off is less than the sample mean

Respuesta :

Answer:

c. Yes because the cut-off is more than the sample mean

Step-by-step explanation:

The null and alternate hypotheses are

H0 : u= 3.5 ounces  vs Ha: u > 3.5 ounces

The mean and standard deviation of the sample are calculated to be 3.76 and 0.24

The test statistics is

t= x`- u/s√n

= 3.76-3.5/0.24/√10

= 0.26/0.07589

=3.426

The critical region for z test is  z > ± 2.33

The critical region for t test is   t > ± 2.82 for 9 d.f

For both the tests( z and t)   the test statistic =3.426 lies  in the critical region.

Hence we reject H0 and accept the hypothesis that  the new design is better than the old one at a level of significance of 1%.

cut-off = mean + 3 times the standard deviation

          =  3.76 + 3* 0.24

 cut-off     =4.48

Yes, the cut off is more than the sample mean.

So choice c is the best option.

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