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Chapter 8: Q102S (page 452)

Find the numerical value of

a.6! b.(109)c. (101)d.(63)e.0!

Short Answer

Expert verified

The results for the given numerical values are as follows:

  1. 720
  2. 10
  3. 10
  4. 20
  5. 1

Step by step solution

01

Important formula

The formula for combination is Crn=n!r!(n-r)!.

02

Finding the value of 6!

a.

6!=6×5×4×3×2×16!=720.

Therefore, the value of 6! is 720.

03

Step 3:Finding the value of (109)

b.

C1910=10!9!10-9!=10!9!1!=10

Accordingly, the value is 10.

04

Evaluate the value of (101)

c.

C110=10!1!10-1!=10!9!1!=10

So, the required value is 10.

05

Finding the value of (63)

d.

C36=6!3!6-3!=6!3!3!=20

Hence, the value is 20.

06

Finding the value of 0!

The value of 0! is 1.

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Most popular questions from this chapter

The data for a random sample of 10 paired observations is shown below.

PairSample from Population 1

(Observation 1)

Sample from Population 2 (Observation 2)
12345678910
19253152493459471751
24273653553466512055

a. If you wish to test whether these data are sufficient to indicate that the mean for population 2 is larger than that for population 1, what are the appropriate null and alternative hypotheses? Define any symbols you use.

b. Conduct the test, part a, usingα=.10.

c. Find a 90%confidence interval for μd. Interpret this result.

d. What assumptions are necessary to ensure the validity of this analysis?

Hospital work-related injuries. According to an Occupational and Health Safety Administration (OHSA) 2014 report, a hospital is one of the most dangerous places to work. The major cause of injuries that resulted in missed work was overexertion. Almost half (48%) of the injuries that result in missed work were due to overexertion. Let x be the number of hospital-related injuries caused by overexertion.

a. Explain why x is approximately a binomial random variable.

b. Use the OHSA report to estimate p for the binomial random variable of part a.

c. Consider a random sample of 100 hospital workers who missed work due to an on-the-job injury. Use the p from part b to find the mean and standard deviation of, the proportion of the sampled workers who missed work due to overexertion.

d. Refer to part c. Find the probability that the sample proportion is less than .40.

Question: Independent random samples from approximately normal populations produced the results shown below.

Sample 1

Sample 2

52 33 42 4441 50 44 5145 38 37 4044 50 43

52 43 47 5662 53 61 5056 52 53 6050 48 60 55

a. Do the data provide sufficient evidence to conclude that (μ1-μ2)>10? Test usingα=0.1.

b. Construct a confidence interval for (μ1-μ2). Interpret your result.

Consider the discrete probability distribution shown here.

x

10

12

18

20

p

.2

.3

.1

.4

a. Calculateμ,σ2 andσ .

b. What isP(x<15) ?

c. Calculate μ±2σ .

d. What is the probability that xis in the interval μ±2σ ?

Question: Consumers’ attitudes toward advertising. The two most common marketing tools used for product advertising are ads on television and ads in a print magazine. Consumers’ attitudes toward television and magazine advertising were investigated in the Journal of Advertising (Vol. 42, 2013). In one experiment, each in a sample of 159 college students were asked to rate both the television and the magazine marketing tool on a scale of 1 to 7 points according to whether the tool was a good example of advertising, a typical form of advertising, and a representative form of advertising. Summary statistics for these “typicality” scores are provided in the following table. One objective is to compare the mean ratings of TV and magazine advertisements.

a. The researchers analysed the data using a paired samples t-test. Explain why this is the most valid method of analysis. Give the null and alternative hypotheses for the test.

b. The researchers reported a paired t-value of 6.96 with an associated p-value of .001 and stated that the “mean difference between television and magazine advertising was statistically significant.” Explain what this means in the context of the hypothesis test.

c. To assess whether the result is “practically significant,” we require a confidence interval for the mean difference. Although this interval was not reported in the article, you can compute it using the information provided in the table. Find a 95% confidence interval for the mean difference and interpret the result. What is your opinion regarding whether the two means are “practically significant.”

Source: H. S. Jin and R. J. Lutz, “The Typicality and Accessibility of Consumer Attitudes Toward Television Advertising: Implications for the Measurement of Attitudes Toward Advertising in General,” Journal of Advertising, Vol. 42, No. 4, 2013 (from Table 1)
See all solutions

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