# 8.5 Additional information and full hypothesis test examples --  (Page 4/51)

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## Try it

A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.

First, determine what type of test this is, set up the hypothesis test, find the p -value, sketch the graph, and state your conclusion.

Since the problem is about percentages, this is a test of single population proportions.

H 0 : p = 0.85

H a : p ≠ 0.85

p = 0.7554

Because p > α , we fail to reject the null hypothesis. There is not sufficient evidence to suggest that the proportion of students that want to go to the zoo is not 85%.

Suppose a consumer group suspects that the proportion of households that have three cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three cell phones.

Set up the Hypothesis Test:

H 0 : p = 0.30 H a : p ≠ 0.30

Determine the distribution needed:

The random variable is P′ = proportion of households that have three cell phones.

The distribution for the hypothesis test is $P\text{'}~N\left(0.30,\sqrt{\frac{\left(0.30\right)\cdot \left(0.70\right)}{150}}\right)$

a. The value that helps determine the p -value is p′ . Calculate p′ .

a. p′ = $\frac{x}{n}$ where x is the number of successes and n is the total number in the sample.

x = 43, n = 150

p′ = $\frac{\text{43}}{\text{150}}$

b. What is a success for this problem?

b. A success is having three cell phones in a household.

c. What is the level of significance?

c. The level of significance is the preset α . Since α is not given, assume that α = 0.05.

d. Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately.
Calculate the p -value.

d. p -value = 0.7216

e. Make a decision. _____________(Reject/Do not reject) H 0 because____________.

e. Assuming that α = 0.05, α < p -value. The decision is do not reject H 0 because there is not sufficient evidence to conclude that the proportion of households that have three cell phones is not 30%.

## Try it

Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the p -value, state your conclusion, and identify the Type I and Type II errors.

H 0 : p = 0.92

H a : p <0.92

p -value = 0.0046

Because p <0.05, we reject the null hypothesis. There is sufficient evidence to conclude that fewer than 92% of American adults own cell phones.

Type I Error: To conclude that fewer than 92% of American adults own cell phones when, in fact, 92% of American adults do own cell phones (reject the null hypothesis when the null hypothesis is true).

Type II Error: To conclude that 92% of American adults own cell phones when, in fact, fewer than 92% of American adults own cell phones (do not reject the null hypothesis when the null hypothesis is false).

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