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# Hypothesis testing

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- In hypothesis testing, a Type 2 error occurs when
- A hypothesis test is done in which the alternative hypothesis is that more than 10% of a population is left-handed. The p-value for the test is calculated to be 0.25. Which statement is correct?
- A significance test based on a small sample may not produce a statistically significant result even if the true value differs substantially from the null value. This type of result is known as
- A result is called “statistically significant” whenever
- Null and alternative hypotheses are statements about:

A. The null hypothesis is not rejected when the null hypothesis is true.

B. The null hypothesis is rejected when the null hypothesis is true.

C. The null hypothesis is not rejected when the alternative hypothesis is true.

D. The null hypothesis is rejected when the alternative hypothesis is true.

A. We can conclude that more than 10% of the population is left-handed.

B. We can conclude that more than 25% of the population is left-handed.

C. We can conclude that exactly 25% of the population is left-handed.

D. We cannot conclude that more than 10% of the population is left-handed.

A. the significance level of the test.

B. the power of the study.

C. a Type 1 error.

D. a Type 2 error.

A. The null hypothesis is true.

B. The alternative hypothesis is true.

C. The p-value is less or equal to the significance level.

D. The p-value is larger than the significance level.

A. population parameters.

B. sample parameters.

C. sample statistics.

D. it depends – sometimes population parameters and sometimes sample statistics

Problem 38

Acme Corporation manufactures light bulbs. The CEO claims that an average Acme light bulb lasts 300 days. A researcher randomly selects 15 bulbs for testing. The sampled bulbs last an average of 290 days, with a standard deviation of 50 days. If the CEO’s claim were true, what is the probability that 15 randomly selected bulbs would have an average life of no more than 290 days?

(A) 0.100

(B) 0.226

(C) 0.334

(D) 0.443

(E) .775

Solution

The answer is (B). The first thing we need to do is compute the t statistic, based on the following equation:

t = [ x – µ ] / [ s / sqrt( n )

]

t = ( 290 – 300 ) / [ 50 / sqrt( 15) ]

t = -10 / 12.909945 = – 0.7745966

where x is the sample mean, µ is the population mean, s is the standard deviation of the sample, and n is the sample size.

Then, using an online calculator (e.g., Stat Trek’s free T Distribution Calculator), a handheldgraphing calculator, or the t distribution table, we find the cumulative probability associated with the t statistic. For this practice test, we can use the T Distribution Calculator; but on the actual AP Statistics Exam, you may need to use a graphing calculator or a t distribution table.

Since we know the t statistic, we select “T score” from the Random Variable dropdown box of the T Distribution Calculator. Then, we enter the following data:

The degrees of freedom are equal to 15 – 1 = 14. The t statistic is equal to – 0.7745966.

The calculator displays the cumulative probability: 0.226. Hence, if the true bulb life were 300 days, there is a 22.6% chance that the average bulb life for 15 randomly selected bulbs would be less than or equal to 290 days.

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