Show so that the expression of cos4A in term of SinA is 8Sin^4 - 8Sin^2A + 1
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Answer:
Claim is rejected
Step-by-step explanation:
Solution:-
- The claim was made on the effectiveness of medication on the market of Parkinson's disease to be p = 75%.
- A random sample was taken of N = 150 individuals and n = 100 number of people reported that it was effectively.
- We are to test the claim made by the manufacturer of the medication based on the statistics available for the sample N.
- State the hypothesis for the effectiveness of medication:
Null Hypothesis: p = 0.75 ... Claim
Alternate hypothesis: p < 0.75 .... Test
- The conditions of standard normality:
- n*p > 5 , 150*0.75 = 112.5 .. ( Check )
- n*(1-p) > 5 , 150*0.25 = 37.5 .. ( Check )
Hence, the standard normal test is applicable. Assuming the population proportion to be normally distributed.
- We will estimate the population proportion with the sample proportion ( p* ):
p* = n / N
p* = 100 / 150
p* = 2/3 = 0.667
- Testing against the claimed population proportion ( p ) = 0.75. The standard normal statistic value is given by:
- We will see whether the Z-test statistic falls in the rejection region defined by the critical value of Z at significance level ( α ) of 0.05.
- The rejection region is defined by the Alternate hypothesis which is less than the claimed value. So, the rejection region defined by the lower tail of the standard normal.
- So for lower tailed test the critical value of statistics is:
P ( Z < Z-critical ) = α = 0.05
Z-critical = - 1.645
- The rejected values all lie to the left of the Z-critical value -1.645
- The claim test value is compared the rejection region:
-2.35607 < -1.645
Z-test < Z-critical
Hence, Null hypothesis rejected because test lies in the rejection region.
Conclusion:
The Null hypothesis or claim made by the manufacturer of Parkinson's disease medication of 75% effectiveness is without sufficient evidence. Hence, the claim made is false or has no statistical evidence.