Question

In San Francisco, a sample of 3,272 wireless routers showed that 1,145 used encryption (to prevent hackers from intercepting information). In Seattle, a sample of 1,940 wireless routers showed that 635 used encryption.

(a) Choose the appropriate hypotheses to test whether or not the population proportion of encryption is higher in San Francisco than Seattle. Assume π₁ is the proportion for San Francisco and π₂ for Seattle.

a. H₀: π₁ − π₂ ≤ 0 vs. H₁: π₁ − π₂ > 0.
b. H₀: π₁ − π₂ ≥ 0 vs. H₁: π₁ − π₂ ≤ 0.

• a

• b

(b-1) Specify the decision rule at α = .05. (Round your answer to 3 decimal places.)

Reject the null hypothesis if z_calc > .____.

(b-2) Find the test statistic z_calc. (Do not round the intermediate calculations and round x₁ and x₂ to the nearest whole number. Round your answer to 3 decimal places.)

z_calc =

(b-3) Choose the correct conclusion.

We reject  Correctthe null hypothesis.

Answer 1

SOLUTION-

a.) We are to test whether the population proportion of encryption is higher in San Francisco than in Seattle.

We already know that denotes the proportion for San Francisco and denotes that of Seattle.

So clearly the second alternative is most suitable for testing the scenario.

b.) WE USE MINITAB 16 FOR CALCULATIONS-

i.) As this is a two sample proportion test with a left tailed alternative, the decision rule is-

At , reject the Null Hypothesis if zcalc. > 1.645

ii.) STEPS- stat> basic statistics> 2 proportions> Enter the summarized values> under "options", set the confidence level-90 and alternative as less than, also use pooled estimate of p for testing> ok.

The test statistic obtained is zcalc = 1.66

iii.) The p-value obtained above is 0.952. As p-value > , we accept the Null Hypothesis and conclude that Population Proportion of Encrypting is more for San Francisco than in Seattle.

REMARK- the necessary explainations have been provided above. In case of any doubt, feel free to comment below and do like if possible.