In: Math
1. A medical school claims that more than 28% of its students plan to go into general practice. It is found that among a random sample of 130 of the school's students, 32% of them plan to go into general practice. Find the P-value for a test of the school's claim
2. In a sample of 47 adults selected randomly from one town, it is found that 9 of them have been exposed to a particular strain of the flu. Find the P-value for a test of the claim that the proportion of all adults in the town that have been exposed to this strain of the flu is 8%.
3. An article in a journal reports that 34% of American fathers take no responsibility for child care. A researcher claims that the figure is higher for fathers in the town of Littleton. A random sample of 225 fathers from Littleton, yielded 97 who did not help with child care. Find the P-value for a test of the researcher's claim
4. An airline claims that the no-show rate for passengers booked on its flights is less than 6%. Of 380 randomly selected reservations, 18 were no-shows. Find the P-value for a test of the airline's claim.
5. Find the P-value for a test of the claim that less than 50% of the people following a particular diet will experience increased energy. Of 100 randomly selected subjects who followed the diet, 47 noticed an increase in their energy level
1) Here we have to test that
n = sample size = 130
Test statistic:
Where
(Round to 4 decimal)
z = 1.02 (Round to 2 decimal)
Test statistic = 1.02
P value:
Test is one tailed(right tailed test)
P value = P(z > 1.02)
= 1 - P(z < 1.02)
= 1 - 0.8461 (From statistical table of z values)
= 0.1539
P value = 0.1539
2) Here we have to test that
n = 47
x = 9
(Round to 4 decimal)
Test statistic:
Where
z = 2.82 (Round to 2 decimal)
Test statistic = 2.82
Test is two tailed test.
P value = 2 * P(z > 2.82)
= 2 * (1- P(z < 2.82))
= 2 * (1 - 0.9976)
= 2 * 0.0024
= 0.0048
P value = 0.0048