Hawaiian Tropic recently ended a marketing and promotion campaign for their new line of sun block...

Hawaiian Tropic recently ended a marketing and promotion campaign for their new line of sun block products containing their patented sunsure technologies. Previous to the campaign 50% of those who use Hawaiian Tropic products used older sun block products. The company surveyed bathers on beaches from Maine to California and found that of a sample of 3,850 users of Hawaiian Tropic products, 2002 were using the new technology sun block products. At the 0.05 level of significance can we conclude that a greater percentage of individuals are using the newer sun block product?

a. State the null and alternative hypotheses. What is the decision rule?

b. What is your decision concerning the null hypothesis? Compute and interpret the p-value. Interpret your decision.

Solutions

Expert Solution

Answer)

Null hypothesis Ho : P = 0.5

Alternate hypothesis Ha : P > 0.5

N = 3850

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 = 1925

N*(1-p) = 1925

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

From z table, P(z>1.645) = 0.05

So, rejection region is

Reject Ho if test statistics is greater than 1.645

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

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

Observed P = 2002/3850

Claimed P = 0.5

N = 3850

After substitution

Test statistics z = 2.48

From z table, P(z>2.48) = 0.0066

P-value = 0.0066

If we will obtain 100 tests assuming null hypothesis is true we will get 0.66 times the result equal to the sample statistics or extreme to the sample statistic.

Since the obtained P-value is less than the given significance level

We reject the null hypothesis Ho

So we have enough evidence to conclude that p > 0.5

orchestra answered 2 years ago

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