Question

Ar esearch manager atCanadaDry claimsthatthe true proportion, p,of CanadaDry drinkers thatpreferCanadaDry overPepsi i s greaterthan0.66. Inaconsumertastetest,170randomly selected people were given blind samples of Canada Dry and Pepsi. 116 of these subjects preferred Canada Dry. (a) Set up the null and alternative hypotheses to test the Canada Dry’s claim. (b) Is there sufficient evidence at the 5% level of significance to validate Canada Dry’s claim? Conduct an appropriate hypothesis test using the p-value method. (c) If your conclusions in parts (b) are wrong, did you commit a type I error or a type II error?

Answer 1

Answer)

A)

Null hypothesis Ho : p = 0.66

Alternate hypothesis Ha : p > 0.66

B)

N = 170

P = 0.66

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 = 112

N*(1-p) = 58

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 = 116/170

Claimed P = 0.66

N = 170

After substitution

Z = 0.62

From z table, P(z>0.62) = 0.2676

P-Value = 0.2676

As the obtained P-Value is greater than the given significance level, we fail to reject Ho

So, we do not have enough evidence to support the claim that p > 0.66

C)

Type 1 error means the rejection of true null hypothesis

Type 2 error means failire to reject false null hypothesis

Now here p = 116/170 = 0.68 which is greater than 0.66

So, technically null hypothesis should be rejected as it is false

But we failed to reject it

So, we have committed a type 2 error