<< Chapter < Page | Chapter >> Page > |
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
$P\text{'}~N\left(0.30,\sqrt{\frac{(0.30)\cdot (0.70)}{150}}\right)$
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%.
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).
Notification Switch
Would you like to follow the 'Statistics i - math1020 - red river college - version 2015 revision a - draft 2015-10-24' conversation and receive update notifications?