Nancy believes that the average running time of movies is equal to 140 minutes. A sample of 4 movies was taken and the following running times were obtained. Assume the distribution of the population is normally distributed. 150 150 180 170 a. State the null and alternative hypotheses.
b. Using a critical value, test the hypothesis at the 10% level of significance.
c. Using a p-value, test the hypothesis at the 10% level of significance.
d. Using a confidence interval, test the hypothesis at the 10% level of significance.
e. Could a Type II error have been committed in this hypothesis test?

a; H0: u= 140 Ha: u ≠ 140 two tailed test.

b. Therefore reject H0 as t= 3 ≠ t≤ t ( ∝/2) (n-1) =2.353 or

3 > t ( ∝/2) (n-1) =2.353    

c. If we check from the table the p- value is 0.6 which lies between 0.1 and 0.05  therefore reject H0.

d. Again reject H0 as 140 < 150.163

e. A type II error has not been committed as H0 is rejected.

Step-by-step explanation:

We formulate the null and alternative hypotheses as

a; H0: u= 140 Ha: u ≠ 140 two tailed test.

For a two tailed test the significance level ∝= 0.1 the critical region is given by

t ≤ t ( ∝/2) (n-1) and t > t ( ∝/2) (n-1)

So the critical region will be

t≤ t ( ∝/2) (n-1) =2.353

where

t= x` - u / s/ √n

Sr. No X X²

1 150 22500

2 150 22500

3 180 32400

4 170 28900

∑ 650 106,300

X`= ∑x/n = 650/4= 162.5

s²= 1/n-1 (x-x`)²= 1/n-1 [ ∑x² -(∑x)²/n ]

= 1/3[106,300 -650²/4] = 225

s= 15

Putting the values in the above equation

t= 162.5- 140/ 15/ √4

t= 3

So calculated value of t= 3

b. Therefore reject H0 as t= 3 ≠ t≤ t ( ∝/2) (n-1) =2.353 or

3 > t ( ∝/2) (n-1) =2.353    

c. If we check from the table the p- value is 0.6 which lies between 0.1 and 0.05  therefore reject H0.

d. a 90% confidence interval based on the calculated values will be

x`± 1.645 (s)/ √n

Putting the values

162.5 ±1.645 ( 15/2)

162.5 ±12.3375

174.84 , 150.163

d. Again reject H0 as 140 < 150.163

e. A type II error has not been committed as H0 is rejected.

step-by-step explanation: