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?
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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?
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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?
Answer)
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
Option 3 is correct answer
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