Question

Question 1

A recent national survey found that high school students watched an average of 6.8 videos per month. A random sample of 36 high school students revealed that the mean number of vidoes watched last month was 6.2. From past experience it is known that the population standard deviation of the number of vidoes watched by high school students is 0.5. At the 0.05 level of signifiance, can we conclude that high school students are watching fewer vidoes?

(a) State the null and alternative hypotheses for this test.

(b) Compute the value of the Test Statistic?

(c) State the p-value for this test.

(d) State the conclusion for the test. Give reasons for your answer.

Question 2

From past records it is known that the average life of a battery used in a digital clock is 305 days. The lives of the batteries are normally distributed. The battery was recently modified to last longer. A sample of 40 modifed batteries was tested. It was discovered that the mean life was 311 days, and the sample standard deviation was 22 days. At the 0.01 level of sigificance, did the modofication increase the mean life of the battery?

(a) State the null and alternative hypotheses for this test.

(b) Compute the value of the Test Statistic?

(c) State the critical region for this test.

(d) State the conclusion for the test. Give reasons for your answer.

Question 3

A machine is set to produce no more than 0.07 defectives when properly adjusted. After the machine had been in operation for some time, a sample of one hundred pieces was tested. Twenty defectives pieces were observed. Is there evidence at the 5% level of significance that the machine needs readjustment?

(a) State the null and alternative hypotheses for this test.

(b) Compute the value of the Test Statistic?

(c) State the p-value for this test.

(d) State the conclusion for the test. Give reasons for your answer.

Answer 1

Question 1

To Test :-

H0 :- µ = 6.8
H1 :- µ < 6.8

Test Statistic :-
Z = ( X - µ ) / ( σ / √(n))
Z = ( 6.2 - 6.8 ) / ( 0.5 / √( 36 ))
Z = -7.2

P value = P ( Z < 7.2 )  = 0 i.e < 0.01

Decision based on P value
Reject null hypothesis if P value < α = 0.05 level of significance
Since 0 < 0.05 ,hence we reject null hypothesis
Result :- Reject null hypothesis

Question 2

To Test :-

H0 :- µ = 305
H1 :- µ > 305

Test Statistic :-
t = ( X̅ - µ ) / (S / √(n) )
t = ( 311 - 305 ) / ( 22 / √(40) )
t = 1.7249

Test Criteria :-
Reject null hypothesis if t > t(α, n-1)
Critical value t(α, n-1) = t(0.01 , 40-1) = 2.426
t > t(α, n-1) = 1.7249 < 2.426
Result :- Fail to reject null hypothesis

Question 3

To Test :-

H0 :- P <= 0.07
H1 :- P > 0.07

Test Statistic :-
Z = ( P - P0 ) / √ ((P0 * q0)/n))
Z = ( 0.2 - 0.07 ) / √(( 0.07 * 0.93) /100))
Z = 5.0951

Decision based on P value
P value = P ( Z > 5.0951 ) = 0 i.e < 0.01
Reject null hypothesis if P value < α = 0.05
Since P value = 0 < 0.05, hence we reject the null hypothesis
Conclusion :- We Reject H0