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In Exercises 13鈥16, refer to the exercise identified and find the value of the test statistic. (Refer to Table 8-2 on page 362 to select the correct expression for evaluating the test statistic.)

Exercise 6 鈥淐ell Phone鈥

Short Answer

Expert verified

The value of the test statistic (z-score) is equal to -12.33.

Step by step solution

01

Given information

Out of 1128 adults, 87% said they have a cell phone.

02

Hypotheses

It is claimed that less than 95% of adults have a cell phone.

Corresponding to the given claim, the following hypotheses are set up:

Null hypothesis: The proportion of adults who have a cell phone is equal to 0.95.

H0:p=0.95

Alternative hypothesis: The proportion of adults who have a cell phone is less than 0.95.

H1:p<0.95

03

Test statistic

Since the claim involves testing the equality of the sample proportion with a hypothesized value, the test statistic used will be the z-score.

The value of the sample proportion is computed below:

p^=87%=87100=0.87

The given value of the proportion ofadults who have cell phones is supposed to be equal to 0.95.

Thus, p=0.95.

q=1-p=1-0.95=0.05

The value of the test statistic is computed below:

z=p^-ppqn=0.87-0.950.950.051128=-12.33

Thus, the test statistic is equal to -12.33.

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Most popular questions from this chapter

Confidence interval Assume that we will use the sample data from Exercise 1 鈥淰ideo Games鈥 with a 0.05 significance level in a test of the claim that the population mean is greater than 90 sec. If we want to construct a confidence interval to be used for testing the claim, what confidence level should be used for the confidence interval? If the confidence interval is found to be 21.1 sec < < 191.4 sec, what should we conclude about the claim?

In Exercises 9鈥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 8 鈥淧ulse Rates鈥

P-Values. In Exercises 17鈥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 = 0.05, should we reject H0or should we fail to reject H0?

The test statistic of z = -2.50 is obtained when testing the claim that p<0.75

Critical Values. In Exercises 21鈥24, refer to the information in the given exercise and do the following.

a. Find the critical value(s).

b. Using a significance level of = 0.05, should we reject H0or should we fail to reject H0?

Exercise 20

In Exercises 1鈥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: 鈥淪hould Americans replace passwords with biometric security (fingerprints, etc)?鈥 Among the respondents, 53% said 鈥測es.鈥 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: 鈥淚f the P is low, the null must go.鈥 What does this mean?

c. Another memory trick commonly used is this: 鈥淚f the P is high, the null will fly.鈥 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?

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