P-Values. In Exercises 17–20, do the following:

a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.

b. Find the P-value. (See Figure 8-3 on page 364.)

c. Using a significance level of α= 0.05, should we reject or should we fail to reject ?

The test statistic of z = 1.00 is obtained when testing the claim that p>0.3.

Short Answer

Expert verified

a. The test is right-tailed.

b. The p-value is equal to 0.1587.

c. The decision of the statistical test is to fail to rejectH0

Step by step solution

01

Given Information

A test statistic value of z=1.00 is obtained, and the claim to be tested is p>0.3.

02

Identify the hypotheses and tail of the test

a.

In correspondence with the given claim, the following hypotheses are set up:

Null Hypothesis: p=0.3

Alternative Hypothesis: p>0.3

Where p is the population proportion.

Since there is greater than sign in the alternative hypothesis, the test is right-tailed.

03

P-value

b.

The test statistic to test the given claim is the z-value.

The z-value is equal to 1.00.

Using the standard normal table, the corresponding right-tailed p-value for z-score equal to 1.00 is equal to:

Pz>1.00=1-Pz<1.00=1-0.8413=0.1587

Thus, the p-value is equal to 0.1587.

To depict the p-value on the normal probability graph, follow the given steps:

  • Plot a horizontal axis representing the z-score. Also label it as “z-score”.
  • Sketch a bell-shaped curve and draw a vertical dotted line corresponding to the value “0” on the horizontal axis
  • Mark the point “1” on the horizontal axis and then shade the area to the right of the value “1” with blue as shown in the figure.
  • Label the shaded region as “p-value = 0.1587”.

The following plot shows the probability value (p-value) as the shaded area under the normal probability graph:

04

Decision about the test

c.

If the p-value is less than the level of significance, the null hypothesis is rejected; otherwise, not.

Here, the level of significance is equal to 0.05, and the p-value is equal to 0.1587.

Since the p-value is greater than 0.05, so the decision is to fail to reject the null hypothesis.

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Most popular questions from this chapter

Type I and Type II Errors. In Exercises 29–32, provide statements that identify the type I error and the type II error that correspond to the given claim. (Although conclusions are usually expressed in verbal form, the answers here can be expressed with statements that include symbolic expressions such as p = 0.1.).

The proportion of people with blue eyes is equal to 0.35.

Testing Claims About Proportions. In Exercises 9–32, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.

Postponing Death An interesting and popular hypothesis is that individuals can temporarily postpone death to survive a major holiday or important event such as a birthday. In a study, it was found that there were 6062 deaths in the week before Thanksgiving, and 5938 deaths the week after Thanksgiving (based on data from “Holidays, Birthdays, and Postponement of Cancer Death,” by Young and Hade, Journal of the American Medical Association, Vol. 292, No. 24). If people can postpone death until after Thanksgiving, then the proportion of deaths in the week before should be less than 0.5. Use a 0.05 significance level to test the claim that the proportion of deaths in the week before Thanksgiving is less than 0.5. Based on the result, does there appear to be any indication that people can temporarily postpone death to survive the Thanksgiving holiday?

In Exercises 9–12, refer to the exercise identified. Make subjective estimates to decide whether results are significantly low or significantly high, then state a conclusion about the original claim. For example, if the claim is that a coin favours heads and sample results consist of 11 heads in 20 flips, conclude that there is not sufficient evidence to support the claim that the coin favours heads (because it is easy to get 11 heads in 20 flips by chance with a fair coin).

Exercise 6 “Cell Phone”

In Exercises 1–4, use these results from a USA Today survey in which 510 people chose to respond to this question that was posted on the USA Today website: “Should Americans replace passwords with biometric security (fingerprints, etc)?” Among the respondents, 53% said “yes.” We want to test the claim that more than half of the population believes that passwords should be replaced with biometric security.

Requirements and Conclusions

a. Are any of the three requirements violated? Can the methods of this section be used to test the claim?

b. It was stated that we can easily remember how to interpret P-values with this: “If the P is low, the null must go.” What does this mean?

c. Another memory trick commonly used is this: “If the P is high, the null will fly.” Given that a hypothesis test never results in a conclusion of proving or supporting a null hypothesis, how is this memory trick misleading?

d. Common significance levels are 0.01 and 0.05. Why would it be unwise to use a significance level with a number like 0.0483?

Identifying H0and H1. In Exercises 5–8, do the following:

a. Express the original claim in symbolic form.

b. Identify the null and alternative hypotheses.

Cell Phone Claim: Fewer than 95% of adults have a cell phone. In a Marist poll of 1128 adults, 87% said that they have a cell phone.

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