Question

A suggestion is made that the proportion of people who have food allergies and/or sensitivities is...

A suggestion is made that the proportion of people who have food allergies and/or sensitivities is 0.54. You believe that the proportion is actually less than 0.54. The hypotheses for this test are Null Hypothesis: p ≥ 0.54, Alternative Hypothesis: p < 0.54. If you select a random sample of 29 people and 15 have a food allergy and/or sensitivity, what is your test statistic and p-value?

Question 7 options:

 1) Test Statistic: 0.246, P-Value: 0.597
 2) Test Statistic: -0.246, P-Value: 0.597
 3) Test Statistic: -0.246, P-Value: 0.403
 4) Test Statistic: -0.246, P-Value: 0.806
 5) Test Statistic: 0.246, P-Value: 0.403

Question 8 (1 point)

A student at a university wants to determine if the proportion of students that use iPhones is different from 0.41. The hypotheses for this scenario are as follows. Null Hypothesis: p = 0.41, Alternative Hypothesis: p ≠ 0.41. If the student randomly samples 25 other students and finds that 9 of them use iPhones, what is the test statistic and p-value?

Question 8 options:

 1) Test Statistic: 0.508, P-Value: 0.611
 2) Test Statistic: -0.508, P-Value: 0.611
 3) Test Statistic: -0.508, P-Value: 0.306
 4) Test Statistic: -0.508, P-Value: 0.694
 5) Test Statistic: -0.508, P-Value: 0.389

Question)

A suggestion is made that the proportion of people who have food allergies and/or sensitivities is 0.54. You believe that the proportion is actually less than 0.54. The hypotheses for this test are Null Hypothesis: p ≥ 0.54, Alternative Hypothesis: p < 0.54. If you select a random sample of 29 people and 15 have a food allergy and/or sensitivity, what is your test statistic and p-value?

N = 29

P = 0.54

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 15.66

N*(1-p) = 13.34

Both the conditions are met so we can use standard normal z table to estimate the P-Value

Test statistics z = (oberved p - claimed p)/standard error

Standard error = √{claimed p*(1-claimed p)/√n

Observed P = 15/29

Claimed P = 0.54

N = 29

After substitution

Test statistics z = -0.246

From z table, P(z<-0.246) = 0.403

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