Question

Include diagrams and calculations for the following: ​​​​​

A.In a recent random sample of 200 students at a large university who were recently forced to watch their lectures on-line, 65 indicated that they preferred on-line lectures as opposed to the normal in-class lectures. Estimate, with 90% confidence, the true proportion of all students at this large university who prefer on-line lectures. (Include a concluding statement similar to concluding statements from examples already covered in class.)

B.In a random sample of 50 Nevada automobile owners who recently had the oil in their cars changed, 30 asked for synthetic oil. Estimate, with 99% confidence, the population proportion of Nevada automobile owners who preferred synthetic oil over regular oil. (No concluding statement is necessary when answering this question.)

Answer 1

Solution :

Given that,

n = 200

x = 65

Point estimate = sample proportion = = x / n = 55 / 200 = 0.325

1 - = 1 - 0.325 = 0.675

At 90% confidence level

= 1 - 90%

=1 - 0.90 = 0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.645 (((0.325 * 0.675 ) / 200 )

= 0.054

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.325 - 0.054 < p < 0.325 + 0.054

0.271 < p < 0.379

( 0.271 , 0.379 )

The 90% confidence interval for the population proportion p is : - ( 0.271 , 0.379 )

( b )

n = 50

x = 30

Point estimate = sample proportion = = x / n = 30 / 50 = 0.600

1 - = 1 - 0.600 = 0.400

At 99% confidence level

= 1 - 99%

=1 - 0.99 = 0.01

/2 = 0.005

Z/2 = Z0.005 = 2.576

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.576 ((( 0.600 * 0.400) / 50)

= 0.178

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.600 - 0.178 < p < 0.600 + 0.178

0.422 < p < 0.778

( 0.422 , 0.778 )

The 99% confidence interval for the population proportion p is : - ( 0.422 , 0.778 )