Question

In: Math

1. A medical school claims that more than 28% of its students plan to go into...

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

Solutions

Expert Solution

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


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